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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 - Mathematics

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प्रश्न

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 a particular teacher is included?

योग
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उत्तर

The number of teachers = 5

Number of students = 20

The number of ways of selecting 2 teachers from 5 teachers is

= 5C2 ways.

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

= `(5!)/(2! xx 3!)`

= `(5 xx 4 xx 3!)/(2! xx 3!)`

= `(5 xx 4)/(2 xx 1)`

= 10 ways

The number of ways of selecting 3 students from 20 students is

= 20C

= `(20!)/(3! xx (20 - 3)!)`

= `(20!)/(3! xx 17!)`

= `(20 xx 19 xx 18 xx 17!)/(3! xx 17!)`

= `(20 xx 19 xx 18)/(3 xx 2 xx 1)`

= 20 × 19 × 3

= 1140 ways

∴ The total number of selection of the committees with 2 teachers and 5 students is

= 10 × 1140

= 11400

A particular teacher is included.

Given a particular teacher is selected.

Therefore, the remaining 1 teacher is selected from the remaining 4 teachers.

Therefore, the number of ways of selecting 1 teacher from the remaining 4 teacher

= 4C1 ways

= 4 ways

The number of ways of selecting 3 students from 20 students = 20C

= `(20!)/(3!(20 - 3)!)`

= `(20!)/(3! xx 17!)`

= `(20 xx 19 xx 18  xx 17!)/(3! xx 17!)`

= `(20 xx 19 xx 18)/(3!)`

= `(20 xx 19 xx 18)/(3 xx 2 xx 1)`

Hence the required number of committees

= 1 × 4 × 1140

= 4560

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Combinations
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 4: Combinatorics and Mathematical Induction - Exercise 4.3 [पृष्ठ १८७]

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सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 4 Combinatorics and Mathematical Induction
Exercise 4.3 | Q 14. (i) | पृष्ठ १८७

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