Question 1

How many 7 digit numbers can be formed using the digits 1, 2, 0, 2, 4, 2, 4?

Accepted Answer

There are 7 digits 1, 2, 0, 2, 4, 2, 4 in which 2 occurs 3 times, 4 occurs 2 times. Number of 7 digit numbers = 7 ! 3 ! × 2 ! = 420 But out of these 420 numbers, there are some numbers which begin with '0' and they are not 7-digit numbers. The number of such numbers beginning with '0'. = 6 ! 3 ! × 2 ! = 60 Hence the required number of 7 digits numbers = 420 - 60 = 360

Question 2

The Indian Cricket team consists of 16 players. It includes 2 wicket keepers and 5 bowlers. In how many ways can a cricket eleven be selected if we have to select 1 wicket keeper and atleast 4 bowlers?

Accepted Answer

We are to choose 11 players including 1 wicket keeper and 4 bowlers or, 1 wicket keeper and 5 bowlers. Number of ways of selecting 1 wicket keeper, 4 bowlers and 6 other players in 2 C 1 * 5 C 4 * 9 C 6 = 840 Number of ways of selecting 1 wicket keeper, 5 bowlers and 5 other players in 2 C 1 * 5 C 5 * 9 C 5 =252 Total number of ways of selecting the team = 840 + 252 = 1092

Question 3

In how many ways can 5 different toys be packed in 3 identical boxes such that no box is empty, if any of the boxes may hold all of the toys ?

Accepted Answer

The toys are different; The boxes are identical If none of the boxes is to remain empty, then we can pack the toys in one of the following ways a. 2, 2, 1 b. 3, 1, 1 Case a. Number of ways of achieving the first option 2 - 2 - 1 Two toys out of the 5 can be selected in 5 C 2 ways. Another 2 out of the remaining 3 can be selected in 3 C 2 ways and the last toy can be selected in 1 C 1 way. However, as the boxes are identical, the two different ways of selecting which box holds the first two toys and which one holds the second set of two toys will look the same. Hence, we need to divide the result by 2 Therefore, total number of ways of achieving the 2 - 2 - 1 option is ways 5 C 2 * 3 C 2 = 15 ways Case b. Number of ways of achieving the second option 3 - 1 - 1 Three toys out of the 5 can be selected in 5 C 3 ways. As the boxes are identical, the remaining two toys can go into the two identical looking boxes in only one way. Therefore, total number of ways of getting the 3 - 1 - 1 option is 5 C 3 = 10 = 10 ways. Total ways in which the 5 toys can be packed in 3 identical boxes = number of ways of achieving Case a + number of ways of achieving Case b= 15 + 10 = 25 ways.

Question 4

How many parallelograms will be formed if 7 parallel horizontal lines intersect 6 parallel vertical lines?

Accepted Answer

Parallelograms are formed when any two pairs of parallel lines (where each pair is not parallel to the other pair) intersect. Hence, the given problem can be considered as selecting pairs of lines from the given 2 sets of parallel lines. Therefore, the total number of parallelograms formed = 7C 2 x 6C 2 = 315

Question 5

The number of ways that 8 beads of different colours be strung as a necklace is

Accepted Answer

The number of ways of arranging n beads in a necklace is ( n - 1 ) ! 2 = ( 8 - 1 ) ! 2 = 7 ! 2 = 2520 (since n = 8)

Question 6

In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?

Accepted Answer

Required number of ways = 8 C 5 * 10 C 6 = 8 C 3 * 10 C 4 = 11760

Question 7

A letter lock consists of three rings each marked with six different letters. The number of distinct unsuccessful attempts to open the lock is at the most ?

Accepted Answer

Since each ring consists of six different letters, the total number of attempts possible with the three rings is = 6 x 6 x 6 = 216. Of these attempts, one of them is a successful attempt. Maximum number of unsuccessful attempts = 216 - 1 = 215.

Question 8

There are 5 multiple choice questions in an examination. How many sequences of answers are possible, if the first three questions have 4 choices each and the next two have 6 choices each?

Accepted Answer

Number of questions = 5 Possibilities of choices for each question 1 to 5 respectively = 4, 4, 4, 6, 6 Reuired total number of sequences = 4 x 4 x 4 x 6 x 6 = 2304.

Question 9

In how many ways can the letters of the word EDUCATION be rearranged so that the relative position of the vowels and consonants remain the same as in the word EDUCATION?

Accepted Answer

The word EDUCATION is a 9 letter word, with none of the letters repeating. The vowels occupy 3rd,5th,7th and 8th position in the word and the remaining 5 positions are occupied by consonants As the relative position of the vowels and consonants in any arrangement should remain the same as in the word EDUCATION, the vowels can occupy only the afore mentioned 4 places and the consonants can occupy1st,2nd,4th,6th and 9th positions. The 4 vowels can be arranged in the 3rd,5th,7th and 8th position in 4! Ways. Similarly, the 5 consonants can be arranged in1st,2nd,4th,6th and 9th position in5! Ways. Hence, the total number of ways = 4! × 5!

Question 10

A decision committee of 5 members is to be formed out of 4 Actors, 3 Directors and 2 Producers. In how many ways a committee of 2 Actors, 2 Directors and 1 Producer can be formed ?

Accepted Answer

Required Number of ways = 4 C 2 × 3 C 2 × 2 C 1 = 36 = 36

Question 11

A polygon has 44 diagonals, then the number of its sides are ?

Accepted Answer

Let the number of sides be n. The number of diagonals is given by nC2 - n Therefore, nC2 - n = 44, n>0 nC2 - n = 44 n2 - 3n - 88 = 0 n2 -11n + 8n - 88 = 0 n(n - 11) + 8(n - 11) = 0 n = -8 or n = 11. As n>0, n will not be -8. Therefore, n=11.

Question 12

Find the total numbers greater than 4000 that can be formed with digits 2, 3, 4, 5, 6 no digit being repeated in any number ?

Accepted Answer

We are having with digits 2, 3, 4, 5 & 6 and numbers greater than 4000 are to be formed, no digit is repeated. The number can be 4 digited but greater than 4000 or 5 digited. Number of 4 digited numbers greater than 4000 are 4 or 5 or 6 can occupy thousand place => 3 x P 3 4 = 3 x 24 = 72. 5 digited numbers = P 5 5 = 5! = 120 So the total numbers greater than 4000 = 72 + 120 = 192

Question 13

How many four digits numbers greater than 6000 can be made using the digits 0, 4, 2, 6 together with repetition.

Accepted Answer

Given digits are 0, 4, 2, 6 Required 4 digit number should be greater than 6000. So, first digit must be 6 only and the remaining three places can be filled by one of all the four digits. This can be done by 1x4x4x4 = 64 Greater than 6000 means 6000 should not be there. Hence, 64 - 1 = 63.

Question 14

Find the number of ways in which 21 balls can be distributed among 3 persons such that each person does not receive less than 5 balls.

Accepted Answer

Let x, y, z be the number of balls received by the three persons, then x ≥ 5 , y ≥ 5 , z ≥ 5 a n d x + y + z = 21 Let u ≥ 0 , v ≥ 0 , w ≥ 0 t h e n ∴ x + y + z =21 ⇒ u + 5 + v + 5 + w + 5 = 21 ⇒ u + v + w = 6 ∴ Total number of solutions = C 3 - 1 6 + 3 - 1 = C 2 8 = 28

Question 15

If nC10=nC12 then,find n.

Accepted Answer

Using, C r n = C n - r n we get n – 10 = 12 or, n = 12 + 10 = 22

Question 16

Four ladies A, B, C and D and four gentlemen E, F, G and H are sitting in a circle round a table facing each other. Directions: (1) No two ladies or two gentlemen are sitting side by side. (2) C, who is sitting between G and E is facing D. (3) F is …

Accepted Answer

G & H , H

Question 17

In How many ways is it possible to make a selection by taking any number of 15 fruits, namely 3 oranges, 5 apples and 7 mangoes?

Accepted Answer

Out of 15 fruits, 7 are alike of one kind, 5 are alike of a second kind and 3 are alike of a third kind. Hence, the required number of ways = [ (7+1) (5+1) (3+1) -1] =191

Question 18

How many different four letter words can be formed (the words need not be meaningful using the letters of the word "MEDITERRANEAN" such that the first letter is E and the last letter is R?

Accepted Answer

The first letter is E and the last one is R. Therefore, one has to find two more letters from the remaining 11 letters. Of the 11 letters, there are 2 Ns, 2Es and 2As and one each of the remaining 5 letters. The second and third positions can either have two different letters or have both the letters to be the same. Case 1: When the two letters are different. One has to choose two different letters from the 8 available different choices. This can be done in 8 * 7 = 56 ways. Case 2: When the two letters are same. There are 3 options - the three can be either Ns or Es or As. Therefore, 3 ways. Total number of possibilities = 56 + 3 = 59

Question 19

There are 7 non-collinear points. How many triangles can be drawn by joining these points?

Accepted Answer

A triangle is formed by joining any three non-collinear points in pairs. There are 7 non-collinear points The number of triangles formed = 7 C 3 = 35

Question 20

There are 6 bowlers and 9 batsmen in a cricket club. In how many ways can a team of 11 be selected so that the team contains at least 4 bowlers?

Accepted Answer

Possibilities Bowlers Batsmen Number of ways 6 9 1 4 7 ( 6 C 4 * 9 C 7 ) 2 5 6 6 C 5 * 9 C 6 3 6 5 6 C 6 * 9 C 5 6 C 4 * 9 C 7 = 15 x 36 = 540 6 C 5 * 9 C 6 = 6 x 84 = 504 6 C 6 * 9 C 5 = 1 x 126 = 126 Total = 1170

Permutation and Combination Questions