SOLVED PROBLEMS ON COMBINATIONS

Problem 1 :

There are 5 teachers and 20 students. Out of them a committee of 2 teachers and 3 students is to be formed. Find the number of ways in which this can be done. Further find in how many of these committees

(i) a particular teacher is included?

(ii) a particular student is excluded?

Solution :

Number of ways in which the commitee can be formed  =  

= ⁵C₂ ⋅  ²⁰C₃

(i) a particular teacher is included?

When a particular teacher always included, it is enough to select the remaining 1 teacher out of 4 teachers.

Total number of ways  = ⁴C₁ ⋅  ²⁰C_3.

(ii) a particular student is excluded?

When a particular student always excluded, it is enough to select the remaining 2 students out of 19 students.

Total number of ways = ⁵C₂ ⋅  ¹⁹C_3.

Problem 2 :

In an examination a student has to answer 5 questions, out of 9 questions in which 2 are compulsory. In how many ways a student can he answer the questions?

Solution :

Total number of questions = 9.

Number of questions must be answered = 2.

Number of questions to be answered = 5.

Number of ways to answer the questions :

= ⁷C₃

  =  35

Problem 3 :

Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly three aces in each combination.

Solution :

Total number of cards in a deck = 52

Number of ace cards = 4

Number of cards to be selected = 5

Here we must select 3 ace cards out of 4 and 2 other cards out of 48.

Number of ways :

= ⁴C₃ ⋅  ⁴⁸C₂

= 4 ⋅ [(48 ⋅ 47)/2]

= 4(1128)

= 4512

Hence the total number of ways is 4512.

Problem 4 :

Find the number of ways of forming a committee of 5 members out of 7 Indians and 5 Americans, so that always Indians will be the majority in the committee.

Solution :

Number of ways = (⁷C₅ ⋅  ⁵C₀) + (⁷C₄ ⋅  ⁵C₁) + (⁷C₃ ⋅  ⁵C₂)

  = (21 ⋅ 1) +  (35 ⋅ 5)  + (35 ⋅ 10)

=  21 + 175 + 350 

= 546

Hence the required number of ways is 546.

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