Q: A group of 9 students, s _1, ~s _2, , s _9, is to be divided to from three teams X , Y , and Z of sizes 2,3 , and 4 , respectively. Suppose that s_1 cannot be selected for the team X , and s _2 cannot

array ccc x & y & z \\ 2 & 3 & 4 array C-i) when x does not contain S _1, but contains S _2 $ (for x) ^7 C_1 (for y,z) 7! 3!4! =245$ C-ii) When x does not contain S _1, ~S _2 and y does not contain S _2 i.e. $ (for x) ^ 7 C _2 (for y,z) 6! 3!3! =420$ so total No. of ways 665

Q: Let S 1 = i , j , k : i , j , k ∈ 1 , 2 , … , 10 S 2 = i , j : 1 ≤ i < j + 2 ≤ 10 , i , j ∈ 1 , 2 , … , 10 S 3 = i , j , k , l : 1 ≤ i

S i = i , j , k : i , j , k ∈ 1 , 2 , ... , 10 n 1 = 10 × 10 × 10 = 1000 i.e., number of elements in S 1 = 1000 S 2 = i , j : 1 ≤ i < j + 2 ≤ 10 , i , j ∈ 1 , 2 , ... , 10 j i 1 1 , 2 2 1 , 2 , 3 ⋮ 8 1 , 2 , 3 , . . . . , 9 n 2 = 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 44 S 3 = i , j , k , l : 1 ≤ i < j < k < l , i , j , k , l ∈ 1 , 2 , ....10 n 3 = 10 C 4 = 10 × 3 × 8 × 7 × 6 ! 4 × 3 × 2

Question 1

The number of ways to distribute 10 identical red pens and 14 identical blue pens among four persons such that each person gets 6 pens, is _____________.

Accepted Answer

Let r_i and b_i be the number of red and blue pens given to the i-th person, where i = 1, 2, 3, 4. Given that each person receives exactly 6 pens, we have: r_i + b_i = 6 b_i = 6 - r_i Since the number of blue pens must be non-negative (b_i 0), we get r_i 6. Also, r_i 0. The total number of red pens is 10, so: r_1 + r_2 + r_3 + r_4 = 10 The number of ways to distribute the pens is the number of non-negative integer solutions to this equation subject to 0 r_i 6. This is equivalent to finding the c

Question 2

A bookshelf contains 6 distinct books of Mathematics and 5 distinct books of Physics. From these 11 books, 6 books are chosen at random. Let X be the absolute value of the difference between the numbe

Accepted Answer

Total number of ways to choose 6 books from 11 books is ^ 11 C_ 6 = 462. Let M and P be the number of Mathematics and Physics books chosen respectively. Since 6 books are chosen, M + P = 6. The random variable X is given by X = |M - P| = |2M - 6|. The possible selections of (M, P), the corresponding values of X, and the number of ways are as follows: For M=1, P=5: X = |1 - 5| = 4, Number of ways = ^ 6 C_ 1 ^ 5 C_ 5 = 6 1 = 6 For M=2, P=4: X = |2 - 4| = 2, Number of ways = ^ 6 C_ 2 ^ 5 C_ 4 = 15

Question 3

Let S be the set of all seven-digit numbers that can be formed using the digits 0,1 and 2 . For example, 2210222 is in S, but 0210222 is NOT in S. Then the number of elements x in S such that at least

Accepted Answer

Let A " 0 " appear exactly twice. and B " 1 " appear exactly twice. A B " 0 " and " 1 " both appears exactly twice. n ( A )=- ------- = _ placing zero ^6 C _2(1)(2)^5= 6 5 2 2^5=480 for n ( B ) C-I : 1 at first place 1 ------- Number of ways = placing 1 ^6 C _1(1)(2)^5=192 C-II : 2 at first place 2 ------- Number of ways = placing 1 ^6 C _2(1)(2)^4 = 6 5 2 2^4=240 n(B)=240+192 for n ( A B ) ------- for n ( A B )= placing zero ^6 C _2(1) placing 1 ^5 C _2(1) 2at rest places 1 (1) = 6 5 2 5 4 2 =1

Question 4

A group of 9 students, s _1, ~s _2, , s _9, is to be divided to from three teams X , Y , and Z of sizes 2,3 , and 4 , respectively. Suppose that s_1 cannot be selected for the team X , and s _2 cannot

Accepted Answer

array ccc x & y & z \ 2 & 3 & 4 array C-i) when x does not contain S _1, but contains S _2 $ (for x) ^7 C_1 (for y,z) 7! 3!4! =245$ C-ii) When x does not contain S _1, ~S _2 and y does not contain S _2 i.e. $ (for x) ^ 7 C _2 (for y,z) 6! 3!3! =420$ so total No. of ways 665

Question 5

Let X  be the set of all five digit numbers formed using 1 , 2 , 2 , 2 , 4 , 4 , 0 . For example, 22240 is in X  while 02244 and 44422 are not in X . Suppose that each element of X  has

Accepted Answer

First we will find the sample space in which the number of five-digit numbers are divisible by 5 , So, fixing zero at the last place we get, - - - - 0 &#9183; Now in first four place following number can take place, 2224 &#160; &#8594; 4 ! 3 ! = 4 &#160; ways 2244 &#8594; 4 ! 2 ! 2 ! = 6 &#160; ways 2221 &#8594; 4 ! 3 ! = 4 &#160; ways 2241 &#8594; 4 ! 2 ! = 12 &#160; ways 2441 &#8594; 4 ! 2 ! = 12 &#160; ways So, total sample space will be  4 + 6 + 4 + 12 + 12 = 38   Now finding the n

Question 6

Consider the 6 &#215; 6  square in the figure. Let A 1 , A 2 , &#8230; , A 49 be the points of intersections (dots in the picture) in some order. We say that A i and A j are friends if they are a

Accepted Answer

Given, We have to choose two distinct points randomly, Now number of ways of selecting 2 adjacent dots in row will be = 6   as if you pick any dot then you have pick second dot from remaining 6  dots, so total ways will be 1 &#215; C 1 6 = 6 Similarly, number of ways of selecting 2 adjacent dots in column = 6 &#8756; &#160; Number of ways of selecting 2 adjacent dots from the given diagram we will   = 6 &#215; 7 + 6 &#215; 7 = 84 So, probability is given by, p = 84 C 2 4

Question 7

Consider 4  boxes, where each box contains 3 red balls and 2 blue balls. Assume that all 20 balls are distinct. In how many different ways can 10 balls be chosen from these 4 boxes so that from e

Accepted Answer

Let the four boxes be Box-1, Box-2, Box-3 and Box-4. Each of the box has  3  red balls and  2  blue balls. 10  balls can be chosen in: Case I - When exactly one box gives four balls Case II - When exactly two boxes gives three balls Required number of ways =  Number of ways in Case I  + Number of ways in Case II. = C 1 4 C 1 2 + C 1 3 C 1 3 C 1 2 3 + C 2 4 C 2 3 C 1 2 + C 1 3 C 2 2 2 C 1 3 C 1 2 2 = 4 5 6 3 + 6 3 &#215; 2 + 3 2 6 2 = 4320 + 17496 = 21816

Question 8

The number of 4 -digit integers in the closed interval 2022 , 4482  formed by using the digits 0 , 2 , 3 , 4 , 6 , 7  is ______.

Accepted Answer

Given the number to be formed by using the digits 0 , 2 , 3 , 4 , 6 , 7   Now assuming counting integers starting from  2 Case-I: if zero on  2 nd  place (i)  2 &#160; 0 &#160; 2 &#160; &#160; _ , now here at unit place  5  number can take place except zero as 2020 &#60; 2022 , so  5  possible way. (ii)  2 &#160; 0 &#160; _ &#160; _ , now here at ten's place  4  number 3 , 4 , 6 , 7  can take place  and at unit place

Question 9

Let  S 1 = i , j , k &#160; : i , j , k &#8712; 1 , 2 , &#8230; , 10 S 2 = i , j &#160; : 1 &#8804; i &#60; j + 2 &#8804; 10 , i , j &#8712; 1 , 2 , &#8230; , 10 S 3 = i , j , k , l : 1 &#8804; i

Accepted Answer

S i = i , j , k : i , j , k &#8712; 1 , 2 , ... , 10 n 1 = 10 &#215; 10 &#215; 10 = 1000 i.e., number of elements in  S 1 = 1000 S 2 = i , j : 1 &#8804; i &#60; j + 2 &#8804; 10 , i , j &#8712; 1 , 2 , ... , 10 j i 1 1 , 2 2 1 , 2 , 3 &#8942;   8 1 , 2 , 3 , . . . . , 9 n 2 = 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 44 S 3 = i , j , k , l : 1 &#8804; i &#60; j &#60; k &#60; l , i , j , k , l &#8712; 1 , 2 , ....10 n 3 = 10 C 4 = 10 &#215; 3 &#215; 8 &#215; 7 &#215; 6 ! 4 &#215; 3 &#215; 2

Question 10

An engineer is required to visit a factory for exactly four days during the first 15  days of every month and it is mandatory that no two visits take place on consecutive days. Then the numb

Accepted Answer

Selection of  4  non-consecutive days out of  15  days is same as first remove 4  objects from  15  objects, so remainging objects are  11  having  12  gaps in which these  4  objects can be placed again. So, total number of ways is  C 4 12 = 12 . 11 . 10 . 9 4 . 3 . 2 . 1 = 495 .

Question 11

In a hotel, four rooms are available. Six persons are to be accommodated in these four rooms in such a way that each of these rooms contains at least one person and at most two persons. Then the numbe

Accepted Answer

6  persons can be distributed in  4  rooms in the group of  1 , 1 , 2 , 2  by  6 ! 1 ! 1 ! 2 ! 2 ! 2 ! 2 ! &#215; 4 ! = 720 &#215; 24 16 = 1080 .

Question 12

Five persons A , &#160; B , &#160; C , &#160; D and E are seated in a circular arrangement. If each of them is given a hat of one of the three colours red, blue and green, then the number of ways of d

Accepted Answer

Given that no two persons sitting adjacent, have the hat of same colour, so hats of all colour must be used, also 3 hats of same colour cannot be used as at least two hats of same colour will become consecutive, so the only combination will be 2 , 2 , 1 . Three cases are possible a ) 2 R , 2 B , 1 G b ) 2 B , 2 G , 1 R c ) 2 R , 2 G , 1 B − 3 way Select are person for single colour hat = C 1 5 (let say green) Now remaining 4 hats can be distributed alternatively in two ways B R B R or R B R B To

Permutation Combination — Previous Year Questions & Solutions