Maths perms and combs

[andy62]

A committee of 4 is to be selected from 6 Aussies and 5 Kiwis. If there must be at least 2 Aussies on the committee, how many different commitees can be formed?

If I consider the cases of 2 Aussies +2, 3 Aussies +1 or 4 Aussies separately I get 265.

However If I decide that there are already 2 Aussies on the commttee so all I need to do is to choose 2 people from the remaining 9, I don't get 265.

Where is the flaw in my logic?

[grasscarp]

Because by leaving the same 2 Aussies on the committee you have given a lesser number of permutations that are possible.

[mohill]

Your first answer is correct.
Your second way of reasoning appears correct at first sight, but it contains many repeats amongst the Aussie selection.
Using your second method you might have a situation as follows
To select 2 Aussies, youhave 6 choices for the first pick and then 5 choices for the second pick. 6x5=30
Suppose one of these ways includes 2 Aussies named A and B. Picking A then B gives the same pair as picking B then A So divide the 30 by 2 and you have 15 ways of picking your pair of Aussies.

In one of your final selections you might have a third Aussie, C, plus any one NZer; ie A B and C and 1kiwi.
Now in another selection, reasoning by this method, you might have A and C as your 2 Aussies and then have B chosen as one of the random further picks, together with the same NZer
This is exactly the same selection as the first one.
And the duplications keep on repeating.

It could be solved by this method but you would go spare considering all of the possble duplications and eliminating them.
I hope this rather rambling reply is helpful
Cheers
mohill

[andy62]

Thanks mohill.