+33616 For
a) How many ways are there of choosing 4 from 14
b) How many ways of choosing 4 from 14 less how many ways of choosing 2 from 12
c) How many ways of choosing 4 from 12 plus how many ways of choosing 2 from 12.
(In general: How many ways of choosing m from n is given by: \(\frac{n!}{(n-m)!\times m!}\) )
+118609 Lets see
a) 4 people are chosen from 14
That is just 14 choose 4 = 14C4 = 1001 (it is not really choose, C stands from combinations)
You say you get that?
b) 2 of the candiates do not want to go together.
ok Let's say those 2 are BOTH chosen
Then there will be 2 chosen out of the remaining 12 which is 12C2 = 66
BUT those are the combinations that are NOT allowed so the number of possible combinations will be
14C4 - 12C2 = 1001 - 66 = 935
Does that help?
+118609 Becasue the two that are not allowed to go together have been taken out 14-2 = 12
What Alan and I have said is
"Lets see how many of the 1001 combinations CONTAIN both those 2 people. "
So now we have already chosen those two candidates.
We have 12 candidates left and we have to choose 2 of them 12C2
That is how many possible delegations of 4 that contain the 2 people who do NOT want to be together
That is the number of possible delegations that must be NOT counted
The number left AFTER these ones are NOT counted are removed 14C4 - 12C2
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Lets's look at an easier questions,
We'll chose 2 letters from ABC
3 letters 2 to chose that is 3C2=3 ways AB, AC, BC
BUT A abd B cannot both be chosen
How many of the 3 ways conatin both A and B
there is one left to choose from and we need to chose 1 so there must be 1 combination with both A and B
(3-2)C(3-2) = 1C1 = 1
All the other combinations do NOT conatin both A and B
what is left is 3C2- 1C1 = 3-1 = 2
