P And C

01.

In a seven digit number only 2 and 3 will present. If no 2’s are consecutive, then the number of such numbers is (A) 26 (B) 33 (C) 32 (D) 53 Hint : Key : B Exactly one 2 and six 3’s = 7C1 = 7 Exactly two 2’s and five 3s = 6C2 = 15 Exactly three 2’s and four 3’s = 5C3 = 10 Exactly four 2’s and three 3’s 4C4 = 1 02. In a polygon no three diagonals are concurrent. If the total number of points of intersection of diagonals interior to the polygon be 70, then the number of diagonals of the polygon is (A) 8 (B) 20 (C) 28 (D) 24 Hint : Key : B Since the points of intersection are interior to the polygon, we have select four vertices so that 2 diagonals intersect. ∴nC4 = 70 ⇒ n(n − 1)(n − 2)(n − 3) = 70 × 24 = 5 × 6 × 7 × 8 ∴n = 8 ∴ No. of diagonals = 8C2 − 8 = 20

03.

The number of positive integers not greater then 100 which are not divisible by 2, 3 or 5 is (A) 26 (B) 36 (C) 30 (D) 42 Hint : Key : A The even numbers 2,4,6……...100 are divisisble by 2. They are 50 in number.The odd numbers which are divisible by 3, There are 10 numbers ending in 5which are divisible by 5.The number of multriples of 3 which do not end with 5 are 3,9,21,27,33,39,51,57,63,69,81,87,93,99. There are 14 such numbers. ∴ required number = 100-50-10-14=26. ⎛n⎞ 04. If ⎜ ⎟ represents the combination of n different things taken k at a time then the value of ⎝k ⎠ ⎛100 ⎞ ⎛ 99 ⎞ ⎛ 98 ⎞ ⎛ 3⎞ ⎛ 2 ⎞ ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ..... + ⎜ ⎟ + ⎜ ⎟ = ⎝ 98 ⎠ ⎝ 97 ⎠ ⎝ 96 ⎠ ⎝1⎠ ⎝ 0⎠ A) 15050 B) 101000 C) 151500 D) 166650 Hint : Key D 2 3 100 i.e, coefficient x2 in (1 + x ) + (1 + x ) + ......... + (1 + x )

05. Sum of the proper divisors of 9900 is (A) 33852 B) 23952 C) 23951 D) 33851 Key :C 9900 = 22 × 32 × 52 × 11 Sum of the divisors = (1 + 2 + 22 )(1 + 3 + 32 )(1 + 5 + 52 ) (1 + 11) =33852

Sum of the proper divisors=33852-(1+9900)=23951 06. The number of ways in which 5 ladies and 7 gentlemen can be seated in a round table so that no two ladies sit together is 7 A) (720) 2 C) 7(720) 2 D) 720 B) 7(360) 2 2

07. A factor P of 10000000099. lies between 9000 and 10,000. The sum of its digits is a) 11 b) 13 c) 17 d) 19 Hint : D 10000000099 = x 5 + x − 1 where x = 100 x5 + x − 1 = ( x 2 − x + 1)( x 3 + x 2 − 1) ⇒ 9901 is a factor of P

⎛n⎞ 08. If ⎜ ⎟ represents the combination of n different things taken k at a time then the value of ⎝k ⎠ ⎛100 ⎞ ⎛ 99 ⎞ ⎛ 98 ⎞ ⎛ 3⎞ ⎛ 2 ⎞ ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ..... + ⎜ ⎟ + ⎜ ⎟ = ⎝ 98 ⎠ ⎝ 97 ⎠ ⎝ 96 ⎠ ⎝1⎠ ⎝ 0⎠ a) 15050 b) 101000 c) 151500 d) 166650 2 3 100 2 Hint : i.e, coefficient x in (1 + x ) + (1 + x ) + ......... + (1 + x ) 09. In a polygon of n sides let, N be the number of diagonals. If N − n > 10 , then the least value of n is (A) 7 (B) 8 (C) 9 (D) 10 Hint : N = nC2 − n n(n − 1) N − n > 10 ⇒ − 2n > 10 ⇒ n 2 − 5n − 10 > 0 2 5 ± 65 ∴n should not lie between 2 10. 15 pigeons are suppose to rest in 5 holes . In how many ways they can rest in holes such that each hole should contain at least one pigeon. (A) 16 c5 (B) 14 c4 (C) 10 c5 (D) 9 c6 Key : B Hint : H1 + H 2 + H 3 + H 4 + H 5 = 15

⇒ (10+5−1) c10 = 14 c10 = 14c4

11. The number of positive integers not greater then 100 which are not divisible by 2, 3 or 5 is (A) 26 (B) 36 (C) 30 (D) 425. Hint : A The numbers 2,4,6……...100 are divisible by 2. They are 50 in number. The odd numbers which are divisible by 3, There are 10 numbers ending in 5which are divisible by 5.The number of multiples of 3 which do not end with 5 are 3,9,21,27,33,39,51,57,63,69,81,87,93,99. There are 14 such numbers. ∴required number = 100-50-10-14=26. 12. There are 4 white 8 blue and 9 yellow shirts. The least number of shirts that can be picked at random so that there are 7 shirts of the same colour is A) 16 B) 17 C) 12 D) 36 Hint: B The least no . of shirts = 4 +6 + 6 +either one blue or one yellow = 16 + 1 = 17 13. 20 candidates are sitting at a round table. One has to select 5 of them so that no two of them sitting side by side are selected. Then the number of ways of selecting the candidates is a) 15C6 b) 13C5 c) 14C6 d) 4 ( 14 c4 )

Hint : D use :

n r

n −r −1

Cr −1

⇒

20 5

20 −5 −1

C5 −1 = 4.14 C4

14. There are 12 girls and 15 boys. Out of them we want to have 4 pairs (each pair contain one boy and one girl) for a dance programme. Number of ways this can be done is 15 a). 12 C4 15C4 b). 12 P4 15C4 c). 12 C4 15C4 8 d). 12 C4 C4 4 4 Hint : D First select 4 girls and 4 boys and then make pair of one boy − one girl Tot. no. of ways = 12 C4 .15 C4 . ( 4 C1.4 C1.3 C1.3 C1.2 C1.2 C1.1C1.1C1 )

= 12 C415 C4 .4!.4! 15. Let x1,x 2 ,x3 be integers greater than 1. If x1.x 2 .x3 = 24.35.53 , then the possible number of the ordered triple ( x1, x 2 , x3 ) is a) 3150 b) 2790 c) 2793 d) 958 Hint : C Number of ways = 6 C 27 C25C2 −3 C1 5 C16C14 C1 −3 C24 C05 C06C0 = 2793 16. The number of ways in which 30 coins of one rupee each be given to six persons, so that none of them receives less than 4 rupees is a) 231 b) 462 c) 693 d) 924 11 Hint : B Required number of ways = C5 = 462 . 17. The number of integral solutions of the equation x1. x2 .x3 .x4 .x5 = 2310 are

( A) 55

( B)

6.55

( C ) 16.55

( D)

56

Hint : C x1 x 2 x 3 x 4 x 5 = 2310 = 3′ 7′ 10′ 11′

each of 3, 7, 10, 11 can be distributed at 5 places in 5 ways

5

+ ve integral sols are 5 a)Two are negative and 3 positive then 5 C3 55 ways b) Four are negative and 1 positive then 5 C4 55 ways. ∴ Total no. of ways = 55 (1 +5 C3 + 5 C4 ) = 16.55 18. The possible number of ordered triples (m,n,p) m n p 1 ≤ m ≤ 100, 1 ≤ n ≤ 50, 1 ≤ p ≤ 25 and 2 + 2 + 2 is divisible by 3 is (A) 31250 (B) 30000 (C) 31249 (D) 32150 Hint : A m n p m n p ( 3 − 1) + ( 3 − 1) + ( 3 − 1) = 3α + ( −1) + 3β + ( −1) + 3γ + ( −1)

such

that

= 3 (α + β + γ ) + ( −1) + ( −1) + ( −1) The above is divisible by 3 if all m,n,p are either even or odd Hence possible number of ordered pairs are 50 × 25 × 13 + 50 × 25 × 12 = 31 250 . 19. There are 4 identical red strips , 3 identical blue strips and 2 identical white strips . The number of flags with three strips in order can be formed using these strips are (A) 10 (B) 15 (C) 20 (D) 25 Hint : C No. of required flags. m

n

p

⎛

= 3! coefficient of . Of x3 in ⎜⎜ 1 + x + ⎝

x 2 x3 x 4 ⎞⎛ x 2 x3 ⎞ ⎛ x2 ⎞ 1+ x + 1+ x + + + + ⎟⎜ ⎟ ⎜ ⎟ = 20 2! 3! 4! ⎟⎜ 2! 3! ⎟⎠ ⎜⎝ 2! ⎟⎠ ⎠⎝

20. Find the number of different ways in which 13 distinct objects can be divided into two groups of 5 and 8 (a) 1287 (b) 1286 (c) 1280 (d) 1387 n! Hint : A Use division of 13 objects into two unequal groups of 5 & 8 pq 21. The number ways of a mixed double game can be arranged from amongst 9 couples if no husband and wife play in the same game is (a) 756 (b) 1512 (c) 3024 (d) 3000 Hint : b 9 Men 9 Couples 9Women First selecting two men, removing corresponding wives, from remaining 7 women two can be selected in 7C2 no of ways = 9C2 × 7C2 × 2

22. The letters of the word MIRROR are arranged in all possible ways these words are written as in a dictionary then the rank of word MIRROR will be (a) 23 (b) 24 (c) 25 (d) 26 Hint: According to dictionary : I, M, O, R, R, R MIO → 1 5! Words begin with I → = 20 MIRO → 1 23 3! NIRROR → 1 23. The number of six digit numbers in which digits are ascending order (a) 48 (b) 84 (c) 120 (d) 126 9×8× 7 Hint : C It is equivalent to 9C6 = 9C3 = = 84 6 24. The value of Expression x = nc0 + (n + 1)C1 + (n + 2)C2 + ......(n + r )Cr is (a) (n + r + 1)Cr 25.

(b) nCr +1

(c) nCn−r

(d) None

Hint : A Replace nCo = n +1C0 use nCr + nCr −1 = n +1Cr Two numbers ‘a’ & ‘b’ are chosen from the set of {1,2,3……3n}. In how many ways can these integers be selected such that a 2 − b 2 is divisible by 3 3 3 1 1 a) n ( n + 1) + n 2 b) n ( n − 1) + n 2 c) n ( n + 1) − n 2 d) n ( n − 1) + n 2 2 2 2 2 Hint : b. G1 :3, 6,9.......3n G2 :1, 4, 7....... ( 3n − 2 ) G3 :2,5,8....... ( 3n − 1) a 2 − b 2 = ( a − b )( a + b )

Either a-b is divisible by 3 (or) a + b is divisible by 3 (or) both nc2 + nc2 + nc2 + nc1.nc1

n ( n − 1) + n2 2 The number of distinct rational numbers of the form p/q, where p, q∈ {1, 2,3, 4,5, 6} is a) 23 b) 32 c) 36 d) 28 Hint: a p = 1, q = 1, 2,3, 4,5, 6 ⇒ 6 3

26.

p = 2, q = 1,3, 4,5, 6 ⇒ 3 ⎡⎣Q ( 2, 4 ) , ( 2, 6 ) ⎤⎦ p = 3, q = 1, 2, 4,5, 6 ⇒ 4 ⎡⎣Q ( 3, 6 ) ⎤⎦

27.

p = 4, q = 1,3,5, 6 ⇒ 3 ⎡⎣Q ( 4, 6 ) ⎤⎦ p = 5, q = 1, 2,3, 4, 6 ⇒ 5 p = 6, q = 1,5 ⇒ 2 Let d1, d2, ……, dk be all the divisors of a positive integer n including 1 and n. Suppose 1 1 1 d1 + d2 + … + dk = 72. Then the value of + + ...... + d1 d 2 dk k2 72 72 b) is c) is 72 k n d) cannot be computed from the given information Hint : c 1 1 1 1 1⎡n n n n⎤ + + + ...... + = ⎢ + + + ...... + ⎥ d1 d 2 d 3 d k n ⎣ d1 d 2 d 3 dk ⎦

a) is

Now

n n , ,...... will also be divisor of the number, i.e., d1 d 2

n = dl for same j and l. dj

1 1 1 1 72 + + ...... + = [ d1 + d 2 + ......] = d1 d 2 dk n n There are 10 stations on a circular path. A train has to stop at 3 stations such that no two stations are adjacent. The number of such selections must be a) 50 b) 84 c) 126 d) None of these 10 Hint : a Total selections = C3 = 120 Number of selections in which 3 stations are adjacent = 10 Number of selections in which 2 stations are adjacent = 6 But there are 10 such pairs. ⇒ Total invalid selections = 10 + 6 × 10 = 70 k ( k + 1) Let n and k be positive integers such that n ≥ . The number of solution (x1, x2, … xk), 2 ⎛ 2n − k 2 + k − 2 ⎞ x1 ≥ 1, x2 ≥ 2,..., xk ≥ k , all integers, satisfying x1 + x2 + … + xk = n, is ⎜ m = ⎟. 2 ⎝ ⎠ m m-1 m b) Ck c) Ck-1 d) Zero a) Ck ⇒

28.

29.

Hint : c Put y1 = x1 – 1, y2 = x2 – 2 …, yk = xk – k k ( k + 1) On adding, y1 + y2 + ... + yk = n − etc. 2 30. An n-digit number is a positive integer with exactly n-digits. Nine hundred distinct n-digit numbers are to be formed by using the digit 2, 5 and 7 only. The smallest value of n for which this is possible is a) 6 b) 7 c) 8 d) 9 Hint : b We must have 3n > 900. The least n satisfying this is 7. 01. For a conference 3 countries have sent 2 delegate each while 3 other countries have sent one delegate each. Number of ways they can be seated in a row so that delegates of the same country are not side by side is x, mark all the correct alter natives for x is a) .244x720 b) 3x8!+760x6!-5! C) 9!-6(8!) + 12 (7!)− 8 (6!) d) None Hint : A, C The total no. of ways = 9! −3 C12!81 +3 C2 2!2!7! −3 C3 2!2!2!6! = 244 × 720 02. There are 7 pigeons and 7 pigeon holes. They come out when they are disturbed by a gunshot. If m is the number of possible ways they go into the holes one each so that none went to their natural habitat, mark all possible alternatives for m. a) 1854 b) 7!− 7c1 6!+ 7c 2 5!− 7c 3 4!+ 7c 4 3!− 7c 5 2!+ 7c 6 − 1 c)

7! 7! 7! 7! 7! − + − + − 7c 0 2! 3! 4! 5! 6!

d) None of these

Key : A, B,C 03. If x is the number of 5 digit numbers sum of whose digits is even and y is the number of 5 digit numbers, sum of whose digit is odd then (a) x = y (b) x + y = 90000 (c) x = 45,000 (d) x < y Hint : a, b, c Total no of 5 digit number ⇒ 9 × 10 × 10 × 10 × 10 x + y = 90,000 but x = y ⇒ x = 45,000 04. The number of selection of four letters taken from the word COLLEGE must be (a) 22 (b) 18 4 (c) Coefficient of x in the expansion of (1 + x + x 2 ) 2 (1 + x)3 (d) 32 L′S − 2 Hint :b, c C, O, G → 3 different letters E ′S − 2 1)Four are different letters = 5C4 = 5 2)2 like 2 different 3) 2 alike , 2 alike

2C1 × 4C2 = 12 = 1

=1 18 05. A class has 30 students. The following prizes are to be awarded to the students of this class first and second in mathematics, first and second in physics, first in chemistry and first in biology, If N denotes the number of ways in which this can be done then, (a) 400 | N (b) 600 | N (c) 8100 | N (d) N divisible by 4 distinct prime numbers. Hint : a, b, c, d 30 p2 × 30 p2 × 30 × 30 06. Letters of the word SUDESH can be arranged in

07.

a) 120 ways when two vowels are together b) 180 ways when two vowels occupy in alphabetical order c) 24 ways when vowels and consonants occupy their respective places d) 240 ways when vowels do not occur together Hint : a, b, c, d 5! 6! 4! ( a ) ( 2!) (b) ( c ) ( 2!) ( d ) 360 − 120 2! 2! 2! The total number of positive integers with distinct digits ( in decimal system) must be 10

a) infinite

b)less than

∑10

i

i=1

10

c) equal to

∑10

d)equal to 9 + 9 × 9 + 9 × 9 × 8 + 9 × 9 × 8 × 7 + ...... + 9 × 9 × 8!

i

i =1

Hint : b, d A positive integer having more than 10 digits cannot have all distinct digits. The number of such numbers is finite. Number of numbers having distinct digits. ⇒ 9 + 9 × 9 + 9 × 9 × 8 + ...... + 9 × 9 × 8! < 10 + 102 + 103 + ...... + 1010 08. Eight people enter an elevator. At each of four floors atleast one person leaves the elevator after which elevator is empty. The number of ways in which this is possible, is 4

a)

∑

4

i =0

Ci ( −1) ( 4 − i ) i

4

8

b)

∑

8

i =0

8

Ci ( −1) ( 8 − i ) i

4

8

d) C4 – 1 c) less than 4 Hint : a, c By principle of exclusion and inclusion, the number of ways 4

= 4 − C1 3 + C2 2 − C31 + C4 0 = ∑ 4 Ci ( −1) ( 4 − i ) 8

4

8

4

8

4

8

4

8

i

8

i =0

Since total ways (unconditional) in which 8 person can align at four floors is 48, the number of ways must be less than 48. 01. Assertion (A) : 260 when divided by 7 leaves remainder 1 n Reason (R ): (1 + x ) = n c0 + n c1 x + n c2 x 2 + ... + n cn x n where n ∈ N (A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true and R is not the correct explanation of A (C ) A is true and R is false (D) A is false and R is true. Key : A Hint : 260 = 820 = (1 + 7) 20 = 1 + 7 ( 20 c1 + 20 c2 7 2 + ... + 20 c19 719 ) 02. Assertion (A): If n is the number of positives integers less then 10,000 which are divisible by all the integers from 2 to 10 (including both), then 1 ≤ n < 5 . because Reason (R) :The number which is a multiple of two positive integers m and n is also a multiple of the least common multiple of both m and n (A)Both A and R are true and R is the correct explanation of A (B)Both A and R are true and R is not the correct explanation of A (C) A is true and R is false (D) A is false and R is true.

Hint : The L.C.M of 2,3,4……….10 is 5×7×8×9=2520 Multiples of 2520 which are less than 10,000 are 2520,2520×2,2520×3. Therefore number of numbers less than 10,000 that are divisible by 2,3,4…10 is 3. ∴n=3 and 1<3<5. Statement 1 is also true. 03. Assertion (A) : Highest power of 10 which can divide (100 ) ! is 24 Reason (R) : According Euler’s concept Highest power of any prime number in n ! can be ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ calculated as follows. Highest power of any prime ( P) = ⎢ n ⎥ + ⎢ n2 ⎥ + ⎢ n3 ⎥ + .... ⎣ p⎦ ⎣ p ⎦ ⎣ p ⎦ (A)Both A and R are true and R is the correct explanation of A (B)Both A and R are true and R is not the correct explanation of A (C) A is true and R is false (D) A is false and R is true. Key : A ⎤ + ..... + ⎡100 ⎤ + ⎡100 ⎤ = 97 Highest power of 2 = ⎡100 ⎤ + ⎡100 ⎢⎣ 2 ⎥⎦ ⎢⎣ 22 ⎥⎦ ⎢⎣ 26 ⎥⎦ ⎢⎣ 27 ⎥⎦ ⎤ + ⎡100 ⎤ = 24 Highest power of 5 = = ⎡100 ⎤ + ⎡100 ⎢⎣ 5 ⎥⎦ ⎢⎣ 52 ⎥⎦ ⎢⎣ 53 ⎥⎦ Highest power of 10 =Power of (2 ∩ 5) =24 04. Assertion (A) :20 Identical balls are distributed into 10 boxes in 29 c9 ways Reason (R ) : Number of positive integral solutions of the equation X 1 + X 2 + .... + X n = r are r −1

cr − n (A)Both A and R are true and R is the correct explanation of A (B)Both A and R are true and R is not the correct explanation of A (C) A is true and R is false (D) A is false and R is true Key : B Since both the statements are correct but explanation is meant for distributing the balls in the boxes such that boxes should contain at least one ball. 05. Assertion (A) : If n is the number of positives integers less then 10,000 which are divisible by all the integers from 2 to 10 (including both), then 1 ≤ n < 5 . because Reason (R ): The number which is a multiple of two positive integers m and n is also a multiple of the least common multiple of both m and n (A)Both A and R are true and R is the correct explanation of A (B)Both A and R are true and R is not the correct explanation of A (C) A is true and R is false (D) A is false and R is true Hint : A The L.C.M of 2,3,4……….10 is 5×7×8×9=2520 Multiples of 2520 which are less than 10,000 are 2520,2520×2,2520×3. Therefore number of numbers less than 10,000 that are divisible by 2,3,4…10 is 3. ∴n=3 and 1<3<5. Statement 1 is also true. 06. Assertion (A) : 8 different flags are used to list on 4 flagstaff . If all the flags are used, then it is considered to be a signal. Then the number of ways in which some of the flagstaffs may not have even single flag is 11!. Because

Reason (R ): Number of ways of distributing n different objects to r persons such that n+r −1Pn some person may not get even one. (A)Both A and R are true and R is the correct explanation of A (B)Both A and R are true and R is not the correct explanation of A (C) A is true and R is false (D) A is false and R is true Hint : D 07. Assertion (A) : The number of ways of getting a total of 17 if 4 different dice are cast is 104 . Reason (R ): distribution of n identical items to r different boxes is such a way that each box may be given at least one item is n−1Pr −1 . (A)Both A and R are true and R is the correct explanation of A (B)Both A and R are true and R is not the correct explanation of A (C) A is true and R is false (D) A is false and R is true Key : C 08. Assertion (A) : If 8 letters are placed in 8 addressed envelopes, then the number of ways of putting the letters in envelopes so that 3 letters go into correct envelopes and none of the remaining letters go into correct envelop is 2464. Because Reason (R ):Number of derangements of n elements from their n habitats is 1 1 1 n 1⎞ ⎛ ⎜ 1 − 1! + 2! − 3! + ...... + ( −1) n! ⎟ ⎝ ⎠

(A)Both A and R are true and R is the correct explanation of A (B)Both A and R are true and R is not the correct explanation of A (C) A is true and R is false (D) A is false and R is true 09. Assertion (A) : Let A = {x / x is a prime number and x<30}. Then the number of different rational numbers. Whose numerator and denominator belong to A is 93. p Reason (R ): is a rational number. ∀ q ≠ 0 and p, q ∈ I q (A)Both A and R are true and R is the correct explanation of A (B)Both A and R are true and R is not the correct explanation of A (C) A is true and R is false (D) A is false and R is true Hint : (D) A = {2,3,5, 7,11,13,17,19, 23, 29}

Two different nos. for Nr and Dr. from these can be obtained in 10 P2 = 90 ways and if

p or p

q =1 q If Dr and Nr same No.of ways = 90 + 1 = 91 10. Assertion (A) : The number of selections of four letters taken from the word PARALLEL must be 15 Because −3 Reason (R ): Coefficient of x 4 in the expansion of (1 − x ) is 15 ( x < 1) (A)Both A and R are true and R is the correct explanation of A (B)Both A and R are true and R is not the correct explanation of A (C) A is true and R is false (D) A is false and R is true Hint : D

1' p, 2 ' A, 1R, 3' L, 1E 4 diff : 5c4 = 5 3 alike of 1 kind & 1 diff = 1c1.4c1 = 4 2 alike of 1 kind & 2 diff = 2c1.4c2 = 2.6 = 12 2 alike of 1 kind & 2 diff of 2nd kind = 2c2 = 1 Total = 22 11. Assertion (A) : If f : {1, 2,3, 4,5} → {1, 2,3, 4,5} then the number of onto functions such that f ( i ) ≠ i is 42

Reason (R ): If n things are arranged in row, the number of ways in which they can be den 1 ⎞ ⎛ 1 1 arranged so that no one of them occupies its original place is n !⎜ 1 − + + ....... ( −1) ⎟ n! ⎠ ⎝ 1! 2! (A)Both A and R are true and R is the correct explanation of A (B)Both A and R are true and R is not the correct explanation of A (C) A is true and R is false (D) A is false and R is true Hint : D 12. Assertion(A): Number of ways of distribution of 12 identical balls into 3 identical boxes is 19 Because Reason (R ):Number of ways of distribution of n identical objects among r persons, each one of whom can receive any number of objects is n + r − 1 cr −1 (A)Both A and R are true and R is the correct explanation of A (B)Both A and R are true and R is not the correct explanation of A (C) A is true and R is false (D) A is false and R is true Hint : D Total 12 identical in 3 distinct 12 + 3 − 1C3−1 = 91 ie. ( x + y + z = 91)

Case (i) When each box contains equal number Case (ii) When two boxes contains equal number 2 x + z = 12 ⇒ ( x = 6, z = 0 )( x = 5, z = 2 ) , ( x = 3, z = 6 )

x = y = z = 4 = 1way

( x = 2, z = 8 )( x = 1, z = 10 ) , ( x = 0, z = 12 ) 18 = 6 ways ⎛ 3! ⎞ ⎜ ⎟ ⎝ 2! ⎠ Case (iii) distinct number 72 = 12 Total − (1 + 18 ) = 72 = 3! ∴ Total = 1 + 6 + 12 = 19 13. Assertion(A):: As n → ∞ , the number of ways of arranging n numbered things in a line where none of them occupies its original position must approach 1/e. 1 1 1 1 Reason (R ): e −1 = 1 − + − + − ...∞ 1! 2! 3! 4! (A)Both A and R are true and R is the correct explanation of A 3c2 .6 = 18 ways =

(B)Both A and R are true and R is not the correct explanation of A (C) A is true and R is false (D) A is false and R is true Hint : D A is false since the number of ways Un. for n things n ⎛ 1 1 1 −1) ⎞ ( = n !⎜ 1 − + − + ... + ⎟ ⎜ 1! 2! 3! ⎟ n ! ⎝ ⎠ U 1 1 1 ⇒ n = 1 − + − + ... 1! 2! 3! n! Probability, not the number of ways approach 1/e. x x 2 x3 R is clearly true since e x = 1 + + + + ...∞ 1! 2! 3! 14. Assertion(A): The number of ways of writing 1400 as a product of two positive integers is 12. Reason (R ):1400 is divisible by exactly three prime numbers (A)Both A and R are true and R is the correct explanation of A (B)Both A and R are true and R is not the correct explanation of A (C) A is true and R is false (D) A is false and R is true Hint : B Since 1400 = 23.52.71 ⇒ Number of factors = (3 + 1)(2 + 1)(1 + 1) = 24. 1 Number of ways of expressing 1400 as a product of two numbers = × 24 = 12. But this does 2 not follow from R which is obviously true. Passage :I 5 letters are supposed to address for 5 people whose addresses are mentioned on envelopes. In how many ways letters are dearrange in envelopes. 1. All the 5 letters are dearranged (i.e all the letters are placed in wrongly addressed) in envelopes a) 120 b) 119 c) 44 d) None of these 2. At least three letters are placed in wrongly addressed envelopes a) 109 b) 110 c) 119 d) None of these 3. Exactly one letter is placed in wrongly addressed envelope a) 2 b) 1 c) 6 d) None of these Hint : Passage : II Let A be a set of n elements. Let p1,p2 ,.....pr be r properties associated with one or more elements of A. Let Ai be the set of elements of A which satisfy at least the property pi;. Ai ∩ Aj be the set of elements of A which satisfy at least the properties Pi and Pj . Let sk be the number of elements of A which satisfy at least k of the properties. Then r

s1 = ∑ n ( A i ) ,s2 = ∑ i =1

∑ n ( Ai ∩ A j ) , etx. n ( A1 ∪ A 2 .... ∪ A r ) = s1 − s2 + s3 ..... + ( −1)

1≤i < J≤r

r +1

sr .

01. Out of first 250 natural numbers, the numbers which are divisible by 6 to 10 or 15 is a) 82 b) 74 c) 176 d) 66 02. Out of the first 250 natural numbers, the number of numbers which are neither divisible by 4 nor by 6 nor by 10 is

a) 95 b) 4 c) 159 d) 155 03. Number of ways of selecting 6 cards out of 52 cards so that there is at least one card from each suit is n. Mark all the correct alternatives for n. b) 13C3 × (13 )3 + 6x (13C2 )2 (13 )2 a) 52C6 − 4 × 39C6 c) 52C6 − 4 ( 39C5 ) + 4C2 26C4 − 4C3 13C3

d)

52

C6 − 4

(

39

) (

C6 + 6

26

) (

C6 − 4

13

C6

)

Key : B Passage : III A is a set containing n elements . A subset S1 of A is chosen. The set A is reconstructed by replacing the element of S1. Again a sub set S2 of A is chosen and again the set is reconstructed by the replacing the elements of S2. The number of ways of choosing S1 or S2 where . 01. S1 and S2 have one element common is (A) 3n−1 (B) n.3n−1 (C) 2n−1 (D) n 02. S1 U S2 = A is (A) 3n (B) n.3n (C) 4n (D) 4n−1 03. S1 is a subset of S2 is (A) 4n−1 (B) 3n+1 (C) 4n (D) 3n KEY : B, A, D Passage : IV If there are n A ′ s, n B′ s and n C′ s making 3n letters. Then the number of ways of selecting r letter out of them. 01. When the value of r ∈ [1,n] is (B) r +1C2 (A) r C1 02. When the value of r ∈ [n + 1, 2n + 1] is

(C)

r +2

(A) r + 2 C2 − 3.r −n+1C2 (B) r C2 − 3.r +n−1C2 03. When the value of r ∈ [ 2n + 2,3n ] is

(C)

r +1

(B) (D)

r +2

C2 − 3.r −n +1C2 + 3r C2

r +2

C2 − 3.r −n+1C2 + 3.r − 2n C2

(A) r + 2 C2 − 3.r −n+1C2 +r −2n C2 (C) r + 2 C2 − 3.r +n−1C2 − 3.r C2 Hint : C,A,D

C2

(D) None of these

C2 − 3.r +n−1C3

(D) r C2 − 3.r +n−1C2

Coefficient of xr is (1 + x + x 2 + ..... + xn )

3

Passage : V Let p be a prime number and n be a positive integer then exponent of p in n ! is denoted by ⎡n⎤ ⎡ n ⎤ ⎡ n ⎤ ⎡ n ⎤ EP ( n !) and is given by E p [ n !] = ⎢ ⎥ + ⎢ 2 ⎥ + ⎢ 3 ⎥ .... + ⎢ n ⎥ where p k < n < p k +1 and [x] ⎣ p⎦ ⎣P ⎦ ⎣P ⎦ ⎣p ⎦ denotes the integer part of x 01. The exponent of 7 in 100 c50 is a) 0 b) 1 c) 2 d) 3 02. The number of zeros at the end of 108! Is a) 10 b) 13 c) 25 d) 26 03. The last no zero digit in 20! must be equal to a) 2 b) 4 c) 6 d) 8

Hint: 01. 02.

100! 50!50! E2 (108!) = 104

100C50 =

E7 ( 50!) = 8 E7 (100!) = 16 E7 (100C50 ) = 16 − 2 ( 8 ) = 0 E5 (108!) = 25

No. of zeros at the end of 108! = E10 (108!) = Min (104, 25 ) = 25 03.

E2 ( 20!) = 18

E3 ( 20!) = 8

E5 ( 20!) = 4

E13 ( 20!) = 1

E17 ( 20!) = 1

E19 ( 20!) = 1

E7 ( 20!) = 2 E11 ( 20!) = 1

20! = 218 × 38 × 54 × 7 2 ×111 × 131 × 171 × 14 1 Last non zero digit = 4 Passage : VI Consider all possible permutations of the word ENDEANOEL 01. The number of permutations containing the word ENDEA (a) 5 (b) 2 5 (c) 7 (d) 6 ! 02. The number of permutations in which the letter E occurs in the first and the last position is (a) 5 (b) 21× 5 (c) 2 × 5 (d) 4 × 5 03. The number of permutations in which A, E, O occur in odd places only (a) 5 (b) 6 (c) 7 (d) 2 × 5 Hint : a, b, d 01. Taking ENDEA as 1 unit Total number of letters are 1+4 = 5 Five letters can be arranged in 5 ways 02.

E ....... E Out of 7 letters 2 are alike then no of ways =

7! = 7 × 6 × 5 × 4 × 3 ×1 2

03. There are 1, 3, 5, 7, 9 → five odd places 5 4 This can be × 3 2 Passage : VII A Triangle is called an integer triangle if all the sides are Integers If a, b, c are sides of an Integer Triangle then we can assume that a ≤ b ≤ c (any other permutation will yield same triangle since sum of two sides is greater than the third side If c is fixed a + b will vary from c + 1 to 2c the number of such integer triangles can be formed by finding integer solutions if a + b = c + 1, a + b = c + 2......a + b = 2c 01. The number of integer isosceles or equilateral triangles none of whose sides exceed 4 must be (a) 9 (b) 10 (c) 11 (d) 12 02. The number of integer isosceles or equilateral triangles none of whose sides exceed 2c must be 3c 2 (a) c 2 (b) 2c 2 (c) 3c 2 (d) 2 03. If c is fixed and odd the number of integer isosceles or equilateral triangles whose sides are a, b, c a ≤ b ≤ c must be

2c − 1 2c + 1 3c + 1 3c − 1 (d) (b) (c) 2 2 2 2 Hint : d, c, d 01. If two sides are equal if equal side is unity. Then third side is <2 ( Q Sum of two sides is greater than third side) Third is 1 If equal side is ‘2’ → third side cannot be > 4 ∴ Third side can be 1, 2, 3 If equal side is ‘3’ → third side cannot be > 6 ∴ Third side can be 1, 2, 3, 4 (Q largest side is 4) Similarly Third side can be 1, 2, 3, 4 1 + 3 + 4 + 4 = 12 02. The value of equal side can be vary from 1 to 2e Third side can be <2, <4………..<4c Total numbers 1+3+5+……..+ 2c – 1 +2c+………..+2c = c 2 + 2c 2 = 3c 2 03. If c = 3, the possible sides 3, 3, 3, 2, 2, 3 3, 2, 3 2, 3, 3 Passage : VIII There are ‘n’ intermediate stations on a railway line from one terminus to another. In how many ways can the train stop at 3 of these intermediate stations, if 01. All the three stations are consecutive a) (n + 2) b) (n + 1) c) (n – 1) d) (n – 2) 02. Atleast two of the stations are consecutive a) (n + 2) (n - 1) b) (n – 2) (n – 1) c) (n – 2)2 d) None 03. No two of these stations are consecutive ( n − 2 )( n − 3) d) none a) nc b) ( n − 2 )c c) 3 3 6 Hint : d ( s1 , s2 , s3 ) , ( s2 , s3 , s4 ) ...... ( sn − 2 , sn −1 , sn ) = ( n − 2 ) (a)

2. c (n – 2) ways (n – 1) ways - (n – 2) = ( n – 2)2 2 3. b nc3 − ( n − 2 ) = ( n − 2 )c 3

Passage : IX Consider the network of equally spaced parallel lines (6 horizontal and 9 vertical) shown in the figure. All small squares are of the same size. A shortest route from A to C is defined as a route consisting 8 horizontal steps and 5 vertical steps. Since any shortest route is a typical 13! arrangement of 8H and 5V. The number of shortest route = . Answer the following 5!8! questions : C D R P

A

Q

B

21. The number of shortest routes through the junction P a) 240 b) 216 c) 560 d) None of these 02. The number of shortest routes which go following street PQ must be a) 324 b) 350 c) 512 d) None of these 03. The number of shortest routes which pass through junctions P and R a) 144 b) 240 c) 216 d) None of these Hint: C Number of shortest routes through P = (Number of shortest routes from A to P)(Number of shortest routes from P to C) 5! 8! = × = 560 3!2! 3!5! 02. B 5! 7! Number of shortest routes = × 1× = 350 3!2! 3!4! 03. B 5! 4! 4! × × = 240 3!2! 3!1! 2!2!

Passage : IX The number of non negative integer solutions of the equation r1x1 + r2x2 + … + rnxn = m where x1, x2, … , xn are the variables and ri are either 1 or greater than 1. If ri = 1 for all i = 1, 2, …, n then it is known that the number of non negative integer solutions is m+n-1Cn-1 and number of positive integer solution is m-1Cn-1. If ri > 1 for some i then we can find the number of integer solutions by assigning all possible integer values to xi. Answer the following questions : 01. The number of integer solutions of the equation 3x1 + 5x2 + x3 + x4 = 10 must be a) 33 b) 34 c) 35 d) 36 02. The number of non negative integer solutions of 5x1 + x2 + x3 + x4 = 14 must be a) 15C2 + 10C2 + 5C2 b) 16C2 + 11C2 + 6C2 16 11 6 c) C3 + C3 + C3 d) None of these 03. The number of positive integer solutions to the equation(x1 + x2 + x3)(y1 + y2 + y3 + y4) = 77 must be

a)

9

C2 ×14 C3 +13 C2 ×10 C3

c) 11 C2 ×15 C3 +14 C2 ×11 C3 Hint : d (Note that x + y = x2 m has m + 1 solutions) x1 0 0 1 0 2 0 0 1 1 1 0 2

b)

(

9

C2 ×14 C3 )

2

d) None of these x3 + x4

No. of Solutions

10 5 0 7 2 4

11 6 1 8 3 5

3 0 1 2 Thus there are 36 solutions. Hint : D (Cases like x1 = 2, x2 = 1 are not possible since 3x1 + 5x2 > 10 for x1 = 2, x2 = 1) 02. b x2 + x3 + x4 No. of Solutions x1 16 14 0 C2 11 9 1 C2 6 4 2 C2 The solutions in the third column are written by using m+n-1Cn-1) 03) B Since 77 = 7 × 11 or 11× 7 We have following possibilities. x1 + x2 + x3 = 7, y1 + y2 + y3 + y4 = 11 x1 + x2 + x3 = 11, y1 + y2 + y3 + y4 = 7 Applying the formula m+n-1Cn-1, we get Number of solutions =9 C2 ×14 C3 +13 C2 ×11 C3 1. Match the column -I with column - II Column - I Column-II a) Rank of TOSS if the letters of the p) 216 word are arranged in dictionary order b)The rank of COCHIN of the letters of q) 97 the word are arranged in dictionary order c) How many different NINE digited numbers can be r) 10 formed from the number 223355888 by rearranging its digits so that the odd digits occupy even positions d) A five digited number divisible by 3 is to be formed s) 60 using the numbers 0.1,2,3,4 & 5 without repetitions. The total number of ways this can be done are 02. Match the following Column-I Column-II a. When three Dice are rolled the number p. 54 of possible out comes in which at least one die shows 6 is b. The number of even proper divisors of 1008 is q. 56 c. The number of ways of selecting 10 balls r. 23 from unlimited red, green, white and yellow balls, if selection must include atleast 2 red and 3 yellow balls is …….. ( where balls of the same colour are alike d. The number of + ve integral solutions of the s. 76 equation xyz = 140 is …. Key : A –S, B – R, C – Q, D – P 03. Consider all possible permutations of the letters of the word ENDEANOEL. Match the statements /Expressions in column − I with the statements /Expressions in Column − II

Column − I Column − II a) The number of permutations containing the word ENDEA, is p) 5 ! b) The number of permutations in which the letter E occurs in the first and the last positions, is q) 2 × 5! c) The number of permutations in which none of the letters D,L,N occurs in the last five positions, is r) 7 × 5! d) The number of permutations in which the letters A, E, O occur only in odd positions, is s) 21 × 5! Key : A – P, B – S, C – Q, D – Q