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 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 :
= ^{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.
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