# WORD PROBLEMS ON COMBINATIONS WITH SOLUTIONS

Problem 1 :

Find the total number of subsets of a set with

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

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

Solution :

(i)   4 elements

nC0 + nC1 + nC2 + · · · + nCn = 2n

here n = 4

=  4C0 + 4C1 + 4C2 + 4C3  4C4

= 24

=  16

(ii)   5 elements

nC0 + nC1 + nC2 + · · · + nCn = 2n

here n = 5

= 25

=  32

(iii)   n elements

nC0 + nC1 + nC2 + · · · + nCn = 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

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

(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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