permutations

Is the answer 20%?

Consider anthony n Micheal to be one group. Now to form a subcomittee, you can choose any one of the 4 ppl - 4C1=4.

The number of ways to split into group of 3 is 6C3=20.

Therefore probability that Anthony n Micheal are together is 4/20=1/5=20%.

anantbhatia wrote:Is the answer 20%?

Consider anthony n Micheal to be one group. Now to form a subcomittee, you can choose any one of the 4 ppl - 4C1=4.

The number of ways to split into group of 3 is 6C3=20.

Therefore probability that Anthony n Micheal are together is 4/20=1/5=20%.

yep!.it's 20% according to me

venmic wrote:

i would love to solve this by this way
select 3 people in 6 people in which A and M are also in group of 3 poeple
we will have 4 combination
now select 3 people in 6 people in which or A or M or no one in the group
suppsed A and M are the same people we will have 5!/2!3!=10 then 10*2 =20 ( because there are 2 times to do this for both A and M)
thus 4/20=0.2 or 20%

The Answer is 40%.

If Micheal must be on one subcommittee, then there are two spots still open (out of five left) for Anthony to sit in, in order for Anthony and Micheal to be on the same committee. Therefore, there the percent that include Anthony is 2/5 = 40%.

can we have the OA plz?