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

##### Solution :-

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)

= (10C2) × (20C3)

By using the formula,

nCr = $$\dfrac{ n!}{r!(n – r)!}$$

10C2 × 20C3 = $$\dfrac{ 10!}{2!(10 – 2)!} ×\dfrac{ 20!}{3!(20-3)!}$$

= 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)

9C1 × 20C3

By using the formula,

nCr =$$\dfrac{ n!}{r!(n – r)!}$$

9C1 × 20C=$$\dfrac{9!}{1!(9 – 1)! }× \dfrac{20!}{3!(20-3)!}$$

= 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)

10C2 × 19C2

By using the formula,

nCr = $$\dfrac{ n!}{r!(n – r)!}$$

10C2 × 19C2 = $$\dfrac{ 10!}{2!(10 – 2)!} × \dfrac{19!}{2!(19-2)! }$$

= 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)

10C2 × 19C3

By using the formula,

nCr =$$\dfrac{ n!}{r!(n – r)!}$$

10C2 × 19C3 = $$\dfrac{ 10!}{2!(10 – 2)!} × \dfrac{19!}{3!(19-3)!}$$

= 5×9 × 19×3×17

= 45 × 969

= 43605 ways

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

Answered by Sakshi | 1 year ago

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