There are 10 professors and 20 students out of whom a committee of 2 professors 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 professor is included.

**(ii)** a particular student is included.

**(iii)** a particular student is excluded.

Asked by Aaryan | 1 year ago | 39

Given:

Total number of professor = 10

Total number of students = 20

Number of ways = (choosing 2 professors out of 10 professors) × (choosing 3 students out of 20 students)

= (^{10}C_{2}) × (^{20}C_{3})

By using the formula,

^{n}C_{r} = \( \dfrac{ n!}{r!(n – r)!}\)

^{10}C_{2} × ^{20}C_{3} = \(\dfrac{ 10!}{2!(10 – 2)!} ×\dfrac{ 20!}{3!(20-3)!}\)

= 5×9 × 10×19×6

= 45 × 1140

= 51300 ways

**(i) **a particular professor is included.

Number of ways = (choosing 1 professor out of 9 professors) × (choosing 3 students out of 20 students)

= ^{9}C_{1} × ^{20}C_{3}

By using the formula,

^{n}C_{r} =\( \dfrac{ n!}{r!(n – r)!}\)

^{9}C_{1} × ^{20}C_{3 }=\( \dfrac{9!}{1!(9 – 1)! }× \dfrac{20!}{3!(20-3)!}\)

= 9 × 10×19×6

= 10260 ways

**(ii) **a particular student is included.

Number of ways = (choosing 2 professors out of 10 professors) × (choosing 2 students out of 19 students)

= ^{10}C_{2} × ^{19}C_{2}

By using the formula,

^{n}C_{r} = \( \dfrac{ n!}{r!(n – r)!}\)

^{10}C_{2} × ^{19}C_{2} = \(\dfrac{ 10!}{2!(10 – 2)!} × \dfrac{19!}{2!(19-2)!
}\)

= 5×9 × 19×9

= 45 × 171

= 7695 ways

**(iii) **a particular student is excluded.

Number of ways = (choosing 2 professors out of 10 professors) × (choosing 3 students out of 19 students)

= ^{10}C_{2} × ^{19}C_{3}

By using the formula,

^{n}C_{r} =\( \dfrac{ n!}{r!(n – r)!}\)

^{10}C_{2} × ^{19}C_{3} = \(\dfrac{ 10!}{2!(10 – 2)!} × \dfrac{19!}{3!(19-3)!}\)

= 5×9 × 19×3×17

= 45 × 969

= 43605 ways

The required no. of ways are 51300, 10260, 7695, 43605.

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