Question

How many different committees can be formed from 10 teachers and 49 students if the committee consists of 2 teachers and 4 students?

Answer 1

Given:

Different committees can be formed from 10 teachers and 49 students if the committee consists of 2 teachers and 4 students.

That is;

( 10 choose 2) = C( 10, 2) and ( 49 choose 4) = C(49, 4)

Recall;

$nC_r=\frac{n!}{(n-r)!r!}$

$10C_2=\frac{10!}{(10-2)!2!}=\frac{10\times9\times8!}{8!\times2!}$

$=5\times9=45$

Similarly,

$49C_4=\frac{49!}{(49-4)!4!}=\frac{49!}{45!\times4!}$$=\frac{49\times48\times47\times46\times45!}{45!\times4!}$$=\frac{49\times48\times47\times46}{4\times3\times2\times1}$$=211876$

Now, multiply the two results.

C(10 , 2) x C(49, 4) = 45 x 211876 = 9534420

Hence, it can be form in 9534420 ways.