Question 1

Compute the sum of 4 digit numbers which can be formed with the four digits 1,3,5,7, if each digit is used only once in each arrangement.

Accepted Answer

The number of arrangements of 4 different digits taken 4 at a time is given by 4 P 4 = 4! = 24.All the four digits will occur equal number of times at each of the position,namely ones,tens,hundreds,thousands. Thus,each digit will occur 24/4 = 6 times in each of the position.The sum of digits in one's position will be 6 x (1+3+5+7) = 96.Similar is the case in ten's,hundred's and thousand's places. Therefore,the sum will be 96 + 96 x 10 + 96 x 100 + 96 x 100 = 106656

Question 2

If repetition of the digits is allowed, then the number of even natural numbers having three digits is :

Accepted Answer

In a 3 digit number one’s place can be filled in 5 different ways with (0,2,4,6,8) 10’s place can be filled in 10 different ways 100’s place can be filled in 9 different ways There fore total number of ways = 5X10X9 = 450

Question 3

In how many different ways can the letters of the word 'THERAPY' be arranged so that the vowels never come together?

Accepted Answer

Given word is THERAPY. Number of letters in the given word = 7 These 7 letters can be arranged in 7! ways. Number of vowels in the given word = 2 (E, A) The number of ways of arrangement in which vowels come together is 6! x 2! ways Hence, the required number of ways can the letters of the word 'THERAPY' be arranged so that the vowels never come together = 7! - (6! x 2!) ways = 5040 - 1440 = 3600 ways.

Question 4

In how many ways the letters of the word NUMERICAL can be arranged so that the consonants always occupy the even places ?

Accepted Answer

NUMERICAL has 9 positions in which 2, 4, 6, 8 are even positions. And it contains 5 consonents i.e, N, M, R, C & L. Hence this cannot be done as 5 letters cannot be placed in 4 positions. Therefore, Can't be determined.

Question 5

5 men and 4 women are to be seated in a row so that the women occupy the even places . How many such arrangements are possible?

Accepted Answer

There are total 9 places out of which 4 are even and rest 5 places are odd. 4 women can be arranged at 4 even places in 4! ways. and 5 men can be placed in remaining 5 places in 5! ways. Hence, the required number of permutations = 4! x 5! = 24 x 120 = 2880

Question 6

In how many ways can 5 letters be posted in 4 letter boxes?

Accepted Answer

First letter can be posted in 4 letter boxes in 4 ways. Similarly second letter can be posted in 4 letter boxes in 4 ways and so on. Hence all the 5 letters can be posted in = 4 x 4 x 4 x 4 x 4 = 1024

Question 7

Using numbers from 0 to 9 the number of 5 digit telephone numbers that can be formed is

Accepted Answer

The numbers 0,1,2,3,4,5,6,7,8,9 are 10 in number while preparing telephone numbers any number can be used any number of times. This can be done in 10 5 ways, but '0' is there So, the numbers starting with '0' are to be excluded is 9 4 numbers. Total 5 digit telephone numbers = 10 5 - 9 4 = 3439

Question 8

If you have 6 New Year greeting cards and you want to send them to 4 of your friends, in how many ways can this be done?

Accepted Answer

We have to find number of permutations of 4 objects out of 6 objects. This number is 6 P 4 = 360 Therefore, cards can be sent in 360 ways.

Question 9

A coach must choose five starters from a team of 12 players. How many different ways can the coach choose the starters ?

Accepted Answer

Choose 5 starters from a team of 12 players. Order is not important. 12 C 5 = 729

Question 10

If a+b+c =21 what is the total number of positive integral solutions?

Accepted Answer

Number of positive integral solutions = n - 1 C r - 1 = C 2 20 = 190

Question 11

If two cards are taken one after another without replacing from a pack of 52 cards. What is the probability for the two cards be Ace ?

Accepted Answer

Total Combination of getting a card from 52 cards = C 1 52 Because there is no replacement, so number of cards after getting first card= 51 Now, Combination of getting an another card= C 1 51 Total combination of getting 2 cards from 52 cards without replacement= ( C 1 52 × C 1 51 ) There are total 4 Ace in stack. Combination of getting 1 Ace is = C 1 4 Because there is no replacement, So number of cards after getting first Ace = 3 Combination of getting an another Ace = C 1 3 Total Combination of getting 2 Ace without replacement= C 1 4 × C 1 3 Now,Probability of getting 2 cards which are Ace = C 1 4 × C 1 3 C 1 52 × C 1 51 = 1/221.

Question 12

In how many different ways could couples be picked from 6 men and 9 women ?

Accepted Answer

Number of mens = 8 Number of womens = 5 Different ways could couples be picked = C 1 6 ×   9 C 1 = 9 x 6 = 54 ways.

Question 13

The number of sequences in which 4 players can sing a song, so that the youngest player may not be the last is ?

Accepted Answer

Let 'Y' be the youngest player. The last song can be sung by any of the remaining 3 players. The first 3 players can sing the song in (3!) ways. The required number of ways = 3(3!) = 4320.

Question 14

In how many ways the word 'SCOOTY' can be arranged such that 'S' and 'Y' are always at two ends?

Accepted Answer

Given word is SCOOTY ATQ, Except S & Y number of letters are 4(C 2O's T) Hence, required number of arrangements = 4!/2! x 2! = 4! = 4 x 3 x 2 = 24 ways.

Question 15

A box contains 2 blue balls, 3 green balls and 4 yellow balls. In how many ways can 3 balls be drawn from the box, if at least one green ball is to be included in the draw?

Accepted Answer

Total number of balls = 2 + 3 + 4 = 9 Total number of ways 3 balls can be drawn from 9 = 9C3 No green ball is drawn = 9 - 3 = 6 = 6C3 Required number of ways if atleast one green ball is to be included = Total number of ways - No green ball is drawn = 9C3 - 6C3 = 9x8x7/3x2 - 6x5x4/3x2 = 84 - 20 = 64 ways.

Question 16

Out of 6 green ball, 4 blue ball, in how many ways we selectone or more balls ?

Accepted Answer

7 × 5 = 35 35 – 1 = 34

Question 17

If A1, A2, A3, A4, ..... A10 are speakers for a meeting and A1 always speaks after, A2 then the number of ways they can speak in the meeting is

Accepted Answer

As A1 speaks always after A2, they can speak only in 1st to 9th places and A2 can speak in 2nd to 10 the places only when A1 speaks in 1st place A2 can speak in 9 places the remaining A3, A4, A5,...A10 has no restriction. So, they can speak in 9.8! ways. i.e when A2 speaks in the first place, the number of ways they can speak is 9.8!. When A2 speaks in second place, the number of ways they can speak is 8.8!. When A2 speaks in third place, the number of ways they can speak is 7.8!. When A2 speaks in the ninth place, the number of ways they can speak is 1.8! Therefore,Total Number of ways they can speak = (9+8+7+6+5+4+3+2+1) 8! = 9 2 ( 9 + 1 ) 8 ! = 10!/2

Question 18

In How many ways can the letters of the word 'CAPITAL' be arranged in such a way that all the vowels always come together?

Accepted Answer

CAPITAL = 7 Vowels = 3 (A, I, A) Consonants = (C, P, T, L) 5 letters which can be arranged in 5 P 5 = 5 ! Vowels A,I = 3 ! 2 ! No.of arrangements = 5! x 3 ! 2 ! =360

Question 19

How many arrangements of the letters of the word ‘BENGALI’ can be made if the vowels are never together.

Accepted Answer

There are 7 letters in the word ‘Bengali of these 3 are vowels and 4 consonants. Considering vowels a, e, i as one letter, we can arrange 4+1 letters in 5! ways in each of which vowels are together. These 3 vowels can be arranged among themselves in 3! ways. Total number of words = 5! x 3!= 120 x 6 = 720

Question 20

There are 5 novels and 4 biographies. In how many ways can 4 novels and 2 biographies can be arranged on a shelf ?

Accepted Answer

4 novels can be selected out of 5 in 5 C 4 ways. 2 biographies can be selected out of 4 in 4 C 2 ways. Number of ways of arranging novels and biographies = 5 C 4 * 4 C 2 = 30 After selecting any 6 books (4 novels and 2 biographies) in one of the 30 ways, they can be arranged on the shelf in 6! = 720 ways. By the Counting Principle, the total number of arrangements = 30 x 720 = 21600

Permutation and Combination Questions