WORD PROBLEMS ON COMBINATIONS WITH SOLUTIONS

Problem 1 :

Find the total number of subsets of a set with

[Hint: ⁿC₀ + ⁿC₁ + ⁿC₂ + · · · + ⁿC_n = 2ⁿ]

(i) 4 elements (ii) 5 elements (iii) n elements

Solution :

(i)   4 elements 

ⁿC₀ + ⁿC₁ + ⁿC₂ + · · · + ⁿC_n = 2ⁿ

here n = 4

  =  ⁴C₀ + ⁴C₁ + ⁴C₂ + ⁴C₃  + ⁴C₄

  = 2⁴

  =  16

(ii)   5 elements 

ⁿC₀ + ⁿC₁ + ⁿC₂ + · · · + ⁿC_n = 2ⁿ

here n = 5

    = 2⁵

  =  32

(iii)   n elements 

ⁿC₀ + ⁿC₁ + ⁿC₂ + · · · + ⁿC_n = 2ⁿ

here n = n

    = 2ⁿ elements 

Problem 2 :

A trust has 25 members.

(i) How many ways 3 officers can be selected?

(ii) In how many ways can a President, Vice President and a Secretary be selected?

Solution :

(i)  Out of 25 members only 3 officers can be selected.

number of ways of selecting 3 officers  =  ²⁵C₃

  =  25!/22! 3!

  =  (25 ⋅ 24 ⋅ 23)/6

  =  2300

(ii) To select a president, we have 9 options 

to select vice president, we have 8 options

to select secretary, we have 7 options

total number of ways  =  9 ⋅ 8 ⋅ 7  =  504

Hence the answer is 504.

Problem 3 :

How many ways a committee of six persons from 10 persons can be chosen along with a chair person and a secretary?

Solution :

Out of 10 members, we have to select a chair person. So we have 10 options to select a chair person. 9 options to select a secretary.

After selecting a chair person and secretary, we have 8 member. Out of 8, we have to select 4 persons.

Hence the answer is (10 ⋅ 9) ⁸C₄

Problem 4 :

How many different selections of 5 books can be made from 12 different books if,

(i) Two particular books are always selected?

(ii) Two particular books are never selected?

Solution :

(i)  Since two particular books are always selected, we may select remaining 3 books out of 10 books.

¹⁰C₃ =  10!/(7! 3!)  =  (10 ⋅ 9 ⋅ 8)/( 3 ⋅ 2) 

  =  120

Hence the answer is 120. 

(ii)  Since two particular books are never selected, we may select 5 books out of 10 books.

¹⁰C₅  =  10!/(5! 5!)  =  (10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6)/(5 ⋅ 4 ⋅ 3 ⋅ 2) 

  =  252

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