a subcommittee of three people is to be chosen from a committee of 10.

M

markosheehan

New member
Joined
May 12, 2016
Messages
31
a subcommittee of three people is to be chosen from a committee of 10.
(i)how many subcommittees are possible?
(ii)how many subcommittees are possible if two particular people on the committee refuse to work together on the subcommittee?
(iii)if the subcommittee is chosen at random what is the probability that the two people who refuse to work together will both end up on the same subcommittee

i can work out the first question it is 10nCr3 which gives a answer of 120 but i cant work out the rest of the question
 
markosheehan said:
a subcommittee of three people is to be chosen from a committee of 10.

(ii)how many subcommittees are possible if two particular people on the committee refuse to work together on the subcommittee?
Click to expand...

In how many ways can three be chosen from eight (being ten, less the two who refuse to work together)?

If, on the other hand, one of the two problem children is chosen, in how many ways can you choose the other two members? ;)
 
markosheehan said:
a subcommittee of three people is to be chosen from a committee of 10.
(i)how many subcommittees are possible?
(ii)how many subcommittees are possible if two particular people on the committee refuse to work together on the subcommittee?
(iii)if the subcommittee is chosen at random what is the probability that the two people who refuse to work together will both end up on the same subcommittee

i can work out the first question it is 10nCr3 which gives a answer of 120 but i cant work out the rest of the question
Click to expand...

Let's call solution to (i) N1

Let's think about (iii) first

Let's tie up those two (obstinate people) - and we have to chose 1 space from 8 - calculate # of ways you can do that ..................... edited

Let's call that number N3

Then probability of choosing those two together is (N3)/(N1) .............................. edited

Now think about solution to (ii)
 
Last edited by a moderator:
Subhotosh Khan said:
Let's call solution to (i) N1

Let's think about (iii) first

Let's tie up those two (obstinate people) - and we have to chose 2 spaces from 9 - calculate # of ways you can do that

Let's call that number N3

Then probability of choosing those two together is (N1)/(N3) should this be the other way around?

Now think about solution to (ii)
Click to expand...

See comment in red.
 
markosheehan said:
a subcommittee of three people is to be chosen from a committee of 10.
(i)how many subcommittees are possible?
(ii)how many subcommittees are possible if two particular people on the committee refuse to work together on the subcommittee?
(iii)if the subcommittee is chosen at random what is the probability that the two people who refuse to work together will both end up on the same subcommittee
the first question it is \(\displaystyle \color{blue}{N_1=\dbinom{10}{3}}\) which gives a answer of 120 , THAT IS CORRECT
Click to expand...
The rest of this question is trivial if we can understand there are only eight ways to form a committee of three that includes both enemies. We see that neither can be chosen or one and not the other chosen.

ii) \(\displaystyle \large N_1-8\).

iii) \(\displaystyle \dfrac{8}{N_1}\)
 
Last edited by a moderator:
You must log in or register to reply here.
Top