Problems on Permutauions and Combinations

Q.1. In how many different ways can the letters of the word ‘JUDGE’ be arranged in such a way that the vowels always come together?

         a) 160    b) 120    c) 48     d) 124      e) 96

Ans: c

Explanation:

In the word ‘JUDGE’ the vowels are ‘UE’

When vowels are taken together it will be treated as one letter

Then the number ways of arrangement of ‘JDG(UE) = 4!

The two vowels can be arranged in 2 ways i.e. 2!

Therefore, the required number of ways = 4! × 2! = 24 × 2 = 48 ways

Q.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?

         a) 1024    b) 540     c) 5040    d) 2520     e) 400

Ans: c

Explanation:

The word ‘LOGARITHMS’ contain  10 letters

Number of 4- letter words, which can be formed from this 10 letters = ¹⁰P₄ = 10 × 9 × 8 × 7

                                                                                                              = 5040 words

Q.3. In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?

           a) 5470    b) 5040     c) 86400    d) 11760     e) 2660

Ans: d

Explanation:

Required number of ways = ⁸C₅ × ¹⁰C₆

                                         = 8 × 7 × 10×  7 × 3 = 11760

Q. 4. How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9 which are divisible by 5 and none of the digits is repeated?

             a) 5   b) 10     c) 15    d) 20     e) 25

Ans: d

Explanation:

Because the numbers should be divisible by 5, the unit’s digit of each numbers must be 5.

Such number of ways =1

The digit at the 10^th place should be any of the remaining 5 digits = 5ways

Then the 100^th digit should be any one from the remaining 4 digits = 4 ways

Therefore, the total number of numbers formed as such = 1 × 5 × 4 = 20

Q. 5. The value of ⁷⁵P₂ is

            a) 2550   b) 2775    c) 1555    d) 5550     e) None of these

Ans: d

Explanation:

ⁿP_r = n!/(n – r)!

⁷⁵P₂ = 75!/73! = 75 × 74 × 73!/73! = 75 × 74 = 5550

Q. 6. How many word can be formed by using all the letters of the word, 'ALLAHABAD' ?

             a) 2520   b) 3270    c) 3780    d) 1890     e) 7560

Ans: e

Explanation:

In the word 'ALLAHABAD’ there are 4A, 2L, 1H, 1B and 1D = 9

Required number of word = 9!/ 4! × 2! × 1! × 1! × 1! = 9  × 8 × 7 × 6 × 5 × 4!/4! × 2

                                           = 9  × 8 × 7 × 3 × 5 = 7560

Q. 7. A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw?

              a) 71   b) 69   c) 64    d) 56     e) 48

Ans: c

Explanation:

Such a combination will be 1 Black + 2 Non-black or 2 Black + 1 Non-black  or 3 black.

          = (³C₁ × ⁶C₂) + (³C₂ × ⁶C₁) + ³C₃

          = (3 × 6 × 5/2 × 1) + (3 × 2/2 × 1 × 6) + 1

          = 45 + 18 + 1 = 64

Q. 8. There are 6 periods in each working day of a school. In how many ways can one organize 5 subjects such that each subject is allowed at least one period?

             a) 3600   b) 2400   c) 3200    d) 2800     e) None of these

Ans: a

Explanation:

5 periods can be organized in ⁶P₅ ways

Remaining period can be arranged in ⁵P₁ ways

Therefore, the total number of ways = ⁶P₅  × ⁵P₁

                                                          = (6 × 5× 4 × 3 × 2) × 5 = 3600

Q. 9. How many 6 digit telephone numbers can be formed if each number starts with 35 and no digit appears more than once?

          a) 1480   b) 720   c) 360    d) 1420     e) 1680

  

Ans: e

Explanation:

The first two places can be filled in only 1 way using 3 and 5.

Remaining 4 places can be filled with any of the 8 remaining digits in ⁸P₄ ways = 8 × 7 × 6 × 5

                                                                              = 1680

Q. 10. An event manager has ten patterns of chairs and eight patterns of tables. In how many ways can he make a pair of table and chair?

             a) 64   b) 110   c) 100    d) 80     e) 70 

Ans: d

Explanation:

A chair can be arranged in 10 ways

A table can be arranged in 8 ways

Therefore, one chair and one table can be arranged in 10 × 8 = 80 ways.