Question 1

A person invites a party of 10 friends at dinner and place so that 4 are on one round table and 6 on the other round table. Total number of ways in which he can arrange the guests is

Accepted Answer

Total person =10 Two groups are formed containing 4 and 6 persons. Total ways of formation of groups = 10! 4!6! Now, 4 persons can be arranged on round table =(4-1)!=3! and 6 persons can be arranged on round table =(6-1) ! =5! Total ways of arranging the guest = 10! 4!6! 3! 5!= 10! 24

Question 2

How many different nine digit numbers can be formed from the number 223355888 by rearranging its digits so that the odd digits occupy even positions

Accepted Answer

Here we have 4 odd digits (3,3,5,5) and 5 even digits (2, 2, 8, 8, 8). O E O E O O E O where, E even place and O odd place Number of ways odd digits will be place on even places = 4! 2!2! Number of ways even digits will be place on odd places aligned & = 5! 2!3! \ & Total number of ways = 4! 2!2! 5! 2!3! =60 aligned

Question 3

If ^ 22 P_ r+1 : ^ 20 P_ r+2 =11: 52, then r is equal to

Accepted Answer

^ 22 P_ r+1 : ^ 20 P_ r+2 =11: 52 or (21-r)(20-r)(19-r)=52 2 21 (21-r)(20-r)(19-r)=14 13 12 aligned & (21-r)(20-r)(19-r)=(21-7)(20-7)(19 \ & -7) aligned r=7

Question 4

At an election a voter may vote for any number of candidates not exceeding the number to be elected. If 4 candidates are to be elected out of the 12 contested in the election and voter votes for at le

Accepted Answer

Total contested candidates =12 4 candidates are to be elected and voter votes for at least one candidate then total number of ways of selections. = ^ 12 C _1+ ^ 12 C _2+ ^ 12 C _3+ ^ 12 C _4 =12+ 12 11 2 1 + 12 11 10 3 2 1 + 12 11 10 9 4 3 2 1 =12+66+220+495=793

Question 5

The number of arrangements of all digits of 12345 such that at least 3 digits will not come in its position is

Accepted Answer

Total number of ways such that at least 3 digits will not come in its position = ^5 C_3 \ 3!- ^3 C_1 2!+ ^3 C_2 1!- ^3 C_3 0! \ + ^5 C_4 \ 4!- ^4 C_1(3!)+ ^4 C_2(2!)- ^4 C_3(1!)+ ^4 C_4 .(0!)\ + ^5 C_4 \ 5!- ^5 C_1 4!+ ^5 C_7 3!- ^5 C_2 2!+ ^5 C_4 1!- . . ^5 C_5(0!) \ =10(2)+5(9)+(44)=20+45+44=109

Question 6

How many different nine digit numbers can be formed from the number 223355888 by rearranging its digits so that the odd digits occupy even positions?

Accepted Answer

X - X - X - X - X . The four digits 3, 3, 5,5 can be arranged at (-) places in 4 ! 2 ! 2 ! =6 ways. The five digits 2,2,8,8,8 can be arranged at (X) places in 5 ! 2 ! 3 ! ways =10 ways Total no. of arrangements =6 10=60 ways

Question 7

Number of words from the letters of the words BHARAT in which B and H will never come together is

Accepted Answer

There are 6 letters in the word BHARAT, 2 of them are identical. Hence total number of words =6 ! / 2 !=360 Number of words in which B and H come together = 5 ! 2 ! 2 ! =120 final ans =360-120=240

Question 8

The number of different seven-digit numbers that can be written using only the three digit 1 , 2 and 3 with the condition that the digit 2 occurs twice in each number is

Accepted Answer

Other than 2, remaining five places can be filled by 1 and 3 for each place. The number of ways for five places is 2 2 2 2 2=2^5. For 2 , selecting 2 places out of 7 is ^7 C_2. Hence, the required number of ways is ^7 C_2 2^5.

Question 9

The number of ways of arranging letters of the word HAVANA so that V and N do not appear together is

Accepted Answer

We can arrange the letters H , A , A , A in 4 ! 3 ! =4 ways. If one possible arrangement is X X X X Then, we can arrange V , N at any of the two places marked with O in the following arrangement O O O O O Thus, we can arrange V and N in ^5 P_2=20 ways. Thus, the number of ways in which letters can be arranged is 4 20=80

Question 10

How many numbers lying between 999 and 10000 can be formed with the help of the digits 0,2,3,6,7,8, when the digits are not repeated?

Accepted Answer

The numbers between 999 and 10000 are all 4-digit numbers. The number of 4 - digit numbers formed by digits 0,2,3,6,7,8 is ^ 6 P _ 4 =360. But here those numbers are also involved which begin from 0 . So, we take those numbers as three-digit numbers. Taking initial digit 0 , the number of ways to fill remaining 3 places from five digits 2,3,6,7,8 are ^ 5 P _ 3 =60 So, the required numbers =360-60=300.

Question 11

If 16 identical pencils are distributed among 4 children such that each gets at least 3 pencils. The number of ways of distributing the pencils is

Accepted Answer

Required number of ways = coeff. of x^ 16 in (x^ 3 +x^ 4 +x^ 5 +x^ 6 +x^ 7 )^ 4

Question 12

The number of possible outcomes in a throw of n ordinary dice in which at least one of the dice shows an odd number is

Accepted Answer

Total number of ways =6 6 to n times =6^ n Total number of ways to show only even number =3 3 to n times =3^ n required number of ways =6^ n -3^ n

Question 13

Total number of ways of selecting 2 white squares on a normal chess board if the squares are not from the same row or column is equal to

Accepted Answer

2 white squares can be selected in ^ 32 C _ 2 ways If they belong to some row or same column then 2 (8 ^ 4 C _ 2 ) ways.

Question 14

A person invites a party of 10 friends at dinner and place them so that 4 are on one round table and 6 on the other round table. The number of ways in which he can arrange the guests is

Accepted Answer

Selection of 6 guests =10 C 6 Permutation of 6 on round table =5 ! Permutation of 4 on round table =3 ! Then, total number of arrangements =10 C 6.5 ! .3 ! = (10) ! 6 ! 4 ! 5 ! .3 != (10) ! 24

Permutation Combination — Previous Year Questions & Solutions