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

Problem 5 :

A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has:

(i) no girl?

(ii) at least one boy and one girl?

(iii) at least 3 girls?

Solution :

Total number of members in the group = 4 girls + 7 boys

= 11 members

i) Selecting no girl, which means we have to select only boys.

Out of 7 boys only 5 boys can be chose.

= ⁷C₅

= 7!/(7 - 5)! 5 !

= 7 x 6 x 5!/2! 5!

= 7 x 3

= 21 ways

ii) at least one boy and one girl

=  ⁷C₁ x ⁴C₄ + ⁷C₂ x ⁴C₃ + ⁷C₃ x ⁴C₂ + ⁷C₄ x ⁴C₁

= 7 x 1 + 21 x 4 + 35 x 6 + 35 x 4

= 7 + 84 + 210 + 140

= 441 ways

iii) at least 3 girls

=  ⁴C₃ x ⁷C₂  +  ⁴C₄ x ⁷C₁

= 4 x 21 + 1 x 7

= 84 + 7

= 91 ways

Problem 6 :

Convert the following products into factorials

5 x 6 x 7 x 8 x 9

Solution :

= 5 x 6 x 7 x 8 x 9

To write it as 9!, we need to multiply both numerator and denominator by 4!

= 5 x 6 x 7 x 8 x 9 x 4!/4!

= (9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1)/4!

= 9!/4!

9!/4! is the factorial form of the given.

Problem 7 :

How many words can be formed using all the letters of the word EQUATION so that

(i) all the vowels are together

(ii) consonants occupy the odd places ?

Solution :

Number of letters in the word EQUATION = 8

Number of vowels = 5

Number of consonants = 3

i)

Group of vowels can be put in one block, the remaining three consonants can be put in separate boxes. So, total of 4 boxes.

= 4! x 5!

= 24 x 120

= 2880

ii) When consonants occupy odd places, we see total of 8 blocks and they should be filled with vowels and consonants.

Out of 4 odd places, 3 places will be filled with vowel because 3 vowels are there. After filling with vowels, the remaining places are 5.

 Out of 5 places 4 places will be filled with consonant and 1 place will be filled with vowel.

= ⁴P₃ x ⁵P₅

= 24 x 5!/0!

the value of 0! is 1.

= 24 x 120 

= 2880

Problem 8 :

From a class of 25 students 10 are to be chosen for an excursion Party. There are 3 students who decide that either all of them will join or none of them will join. In how many ways can excursion party be chosen?

Solution :

Total number of students = 25

Number of students to be chosen = 10

Since 3 students has decided to join, the remaining 7 students may be chosen.

= ²²C₇

or

Since that the 3 students are deciding not to join, the 10 students should be chosen out of 22 students.

^{= 22}C₁₀

Total number of ways =  ²²C₇ +  ²²C₁₀

= 22!/(22 - 7)!7! + 22!/(22 - 10)! x 10!

= 22!/15!7! + 22!/12! x 10!

22!/15!7!

= (22 x 21 x 20 x 19 x 18 x 17 x 16) / (7 x 6 x 5 x 4 x 3 x 2)

= 170544

22!/12! x 10!

= (22 x 21 x 20 x 19 x 18 x 17 x 16 x 15 x 14 x 13)/(10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2)

= 646646

Total number of ways = 170544 + 646646

= 817190

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.