Answer :

Given that we need to choose 2 professors and 3 students out of 10 professors and 20 students,

Let us assume the choosing the no. of ways be N,

⇒ N = (choosing 2 professors out of 10 professors) × (choosing 3 students out of 20 students)

⇒ N = (^{10}C_{2}) × (^{20}C_{3})

We know that

And also n! = (n)(n – 1)(n – 2)…………2.1

⇒

⇒

⇒

⇒ N = 45 × 1140

⇒ N = 51300 ways

It is told that a particular is always included.

It is similar to selecting 1 professor and 3 students out of the remaining 9 professors and 20 students as 1 professor is already selected.

Let us assume the choosing the no. of ways be N_{1},

⇒ N_{1} = (choosing 1 professor out of 9 professors) × (choosing 3 students out of 20 students)

⇒ N_{1} = ^{9}C_{1} × ^{20}C_{3}

We know that

And also n! = (n)(n – 1)(n – 2)…………2.1

⇒

⇒

⇒

⇒ N_{1} = 9 × 1140 ways

⇒ N_{1} = 10260 ways

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