Solved Problems on Combinations :
In this section, we will learn, how to solve 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 =
= ^{5}C_{2} ⋅ ^{20}C_{3}
(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 = = ^{4}C_{1} ⋅ ^{20}C_{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 = = ^{5}C_{2} ⋅ ^{19}C_{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 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 = 9 - 2 = ^{7}C_{3}
= 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 = ^{4}C_{3 }⋅ ^{48}C_{2}
= 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 :
7 Indians |
5 Americans |
Committee to be formed |
5 4 3 |
0 1 2 |
5 5 5 |
Number of ways = (^{7}C_{5 }⋅ ^{5}C_{0}) + (^{7}C_{4 }⋅ ^{5}C_{1}) + (^{7}C_{3 }⋅ ^{5}C_{2})
= (21 ⋅ 1) + (35 ⋅ 5) + (35 ⋅ 10)
= 21 + 175 + 350
= 546
Hence the required number of ways is 546.
After having gone through the stuff given above, we hope that the students would have understood, how to solve problems on combinations.
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