Question

A bag contains 6 black and 5 red balls. The number of ways in which 3 black balls are selected is m and 2 black balls and 3 red balls selected is n then (m,n)=

Answer 1

Answer:

There are 5 black and 6 red balls in the bag.

2 black balls can be selected out of 5 black ball in

5

C

2

ways and 3 red ball can be selected out of 6 red balls in

6

C

3

ways.

Thus, by multiplication principle, required number of ways of selecting 2 black and 3 red balls is

5

C

2

×

6

C

3

=

2!3!

5!

×

3!3!

6!

=

2

5×4

×

3×2×1

6×5×4

=10×20=200