Question 1

How many arrangements can be made out of the letters of the word COMMITTEE, taken all at a time, such that the four vowels do not come together?

Accepted Answer

There are total 9 letters in the word COMMITTEE in which there are 2M's, 2T's, 2E's. The number of ways in which 9 letters can be arranged = 9 ! 2 ! × 2 ! × 2 ! = 45360 There are 4 vowels O,I,E,E in the given word. If the four vowels always come together, taking them as one letter we have to arrange 5 + 1 = 6 letters which include 2Ms and 2Ts and this be done in 6 ! 2 ! × 2 ! = 180 ways. In which of 180 ways, the 4 vowels O,I,E,E remaining together can be arranged in 4 ! 2 ! = 12 ways. The number of ways in which the four vowels always come together = 180 x 12 = 2160. Hence, the required number of ways in which the four vowels do not come together = 45360 - 2160 = 43200

Question 2

How many 4-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed?

Accepted Answer

'LOGARITHMS' contains 10 different letters. Required number of words = Number of arrangements of 10 letters, taking 4 at a time. = 10 P 4 = 5040.

Question 3

When four fair dice are rolled simultaneously, in how many outcomes will at least one of the dice show 3?

Accepted Answer

When 4 dice are rolled simultaneously, there will be a total of 6 x 6 x 6 x 6 = 1296 outcomes. The number of outcomes in which none of the 4 dice show 3 will be 5 x 5 x 5 x 5 = 625 outcomes. Therefore, the number of outcomes in which at least one die will show 3 = 1296 – 625 = 671

Question 4

In a Plane there are 37 straight lines, of which 13 pass through the point A and 11 pass through the point B. Besides, no three lines pass through one point, no lines passes through both points A and B , and no two are parallel. Find the number of p…

Accepted Answer

The number of points of intersection of 37 lines is C 2 37 . But 13 straight lines out of the given 37 straight lines pass through the same point A. Therefore instead of getting C 2 13 points, we get only one point A. Similarly 11 straight lines out of the given 37 straight lines intersect at point B. Therefore instead of getting C 2 11 points, we get only one point B. Hence the number of intersection points of the lines is C 2 37 - C 2 13 - C 2 11 + 2 = 535

Question 5

How many necklace of 12 beads each can be made from 18 beads of different colours?

Accepted Answer

Here clock-wise and anti-clockwise arrangements are same. Hence total number of circular–permutations: 18 P 12 2 * 12 = 18 ! 6 * 24

Question 6

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 7

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 8

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 9

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 10

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 11

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 12

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 13

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 14

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 15

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 16

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 17

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 18

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 19

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 20

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

Permutation and Combination Questions