PRACTICE WORD PROBLEMS ON COMBINATIONS WITH SOLUTIONS

Problem 1 :

A committee of 7 peoples has to be formed from 8 men and 4 women. In how many ways can this be done when the committee consists of

(i) exactly 3 women?

(ii) at least 3 women?

(iii) at most 3 women?

Solution :

Number of men  =  8

Number of women  =  4

(i) exactly 3 women?

Number of ways  =  ⁴C₃ ⋅  ⁸C₄

=  4 (70)

=  280

(ii) at least 3 women?

Number of ways  =  (⁴C₃ ⋅  ⁸C₄) +  (⁴C₄ ⋅  ⁸C₃)

  =  4(70) + 1(56)

  =  280 + 56

  =  336

(iii) at most 3 women?

Number of ways 

  =  (⁴C₀ ⋅  ⁸C₇) +  (⁴C₁ ⋅  ⁸C₆) +  (⁴C₂ ⋅  ⁸C₅)  +  (⁴C₃ ⋅  ⁸C₄)

  =  8 + 4(28) + 6(56) + 4(70)

  =   8 + 112 + 336 + 280

  =  736

Problem 2 :

7 relatives of a man comprises 4 ladies and 3 gentlemen, his wife also has 7 relatives; 3 of them are ladies and 4 gentlemen. In how many ways can they invite a dinner party of 3 ladies and 3 gentlemen so that there are 3 of man’s relative and 3 of the wife’ s relatives?

Solution :

(i) 3 ladies from husband’s side and 3 gentlemen from wife’s side.

No. of ways in this case

  = ⁴C₃ ⋅ ⁴C₃ = 4 ⋅ 4 = 16

(ii) 3 gentlemen from husband’s side and 3 ladies from wife’s side.

No. of ways in this case = ³C₃ ⋅ ³C₃ = 1 ⋅ 1 = 1

(iii) 2 ladies and one gentleman from husband’s side and lady and 2 gentlemen from wife’s side.

No. of ways in this case

= (⁴C₄ ⋅ ³C₁) ⋅ (³C₁ ⋅ ⁴C₂) = 6 ⋅ 3 ⋅ 3 ⋅ 6 = 324

(iv) One lady and 2 gentlemen from husband’s side and 2 ladies and one gentlemen from wife’s side.

No. of ways in this case

  =  (⁴C₁ ⋅ ³C₂) ⋅ (³C₂ ⋅ ⁴C₁) = 4 ⋅ 3 ⋅ 3 ⋅ 4 = 144

Hence the total no. of ways are

=  16 + 1 + 324 + 144 = 485 ways 

Problem 3 :

Suppose that 7 people enter a swim meet. Assuming that there are no ties, in how many ways could the gold, silver, and bronze medals be awarded?

Solution :

Since the order matters here, we have to use the concept of permutation.

= ⁷P₃

= 7!/(7 - 3)!

= 7!/4!

= 7 x 6 x 5

= 210 ways

Problem 4 :

How many different committees of 3 people can be chosen to work on a special project from a group of 9 people?

Solution :

Number of people in the group = 9

Number of people to be chosen = 3

Here choosing person is not a matter, we use the concept of combination.

= ⁹C₃

= 9!/(9 - 3)! 3!

= 9!/6!3!

= (9 x 8 x 7) / (3 x 2 x 1)

= 84

Problem 5 :

A coach must choose how to line up his five starters from a team of 12 players. How many different ways can the coach choose the starters?

Solution :

The coach has to choose 5 players out of 12 players, the order matters.

= ¹²P₅

= 12!/(12-5)!

= 12 x 11 x 10 x 9 x 8 x 7!/7!

= 95040

Problem 6 :

John bought a machine to make fresh juice. He has five different fruits: strawberries, oranges, apples, pineapples, and lemons. If he only uses two fruits, how many different juice drinks can John make?

Solution :

Number of different fruits available = 5

Number of fruits she can use = 2

= ⁵C₂

= 5!/(5-2)!2!

= (5 x 4 x 3!)/3! 2!

= 20/2

= 10 ways

Problem 7 :

How many different four-letter passwords can be created for a software access if no letter can be used more than once?

Solution :

Total number of alphabets = 26

Out of 26 letters only 4 letters to be chosen, then 

= ²⁶P₄

= 26 x 25 x 24 x 23

= 358800

Problem 8 :

.How many different ways you can elect a Chairman and Co-Chairman of a committee if you have 10 people to choose from.

Solution :

Total number of people in the committee = 10

Two positions are there, Chairman and Co-Chairman

Order matters, so 10P2

= 10!/8!

= 10 x 9

= 90 ways

Problem 9 :

There are 25 people who work in an office together. Five of these people are selected to go together to the same conference in Orlando, Florida. How many ways can they choose this team of five people to go to the conference?

Solution :

Number of people = 25

= ²⁵C₅

= 25!/20!5!

= (25 x 24 x 23 x 22 x 21)/(5 x 4 x 3 x 2 x 1)

= 5 x 23 x 22 x 21

= 53130

Problem 10 :

There are 25 people who work in an office together. Five of these people are selected to attend five different conferences. The first person selected will go to a conference in Hawaii, the second will go to New York, the third will go to San Diego, the fourth will go to Atlanta, and the fifth will go to Nashville. How many such selections are possible?

Solution :

Total number of people = 25

Number of people to be selected = 5

= ²⁵P₅

= 25!/20!

= 25 x 24 x 23 x 22 x 21

= 6375600

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