Maths - Combinations question

[LiquidIce]

I'm repeating my maths materials for the upcoming exams in a few days but I got stuck in the wee chapter called "Combinations"

It's pretty damn simple, nCr and stuff, but one questions stumped me:

A class consists of 12 boys of whom 5 are prefects. How many committees of 8 can be formed it the committee is to have:
a) 3 prefects?
b) at least 3 prefects?

I solved point a) by thinking along these lines:
3 have to be prefects, so those 3 will be chosen from the 5 prefects - 5 nCr 3 = 10. This means that 3 out of 8 have been chosen and now the 5 will be chosen from the non-prefects - 7 boys: 7 nCr 5 = 21.
21 * 10 = 210 combinations which is right according to the answers.

Now this is my problem:
I apply the exact same solving mechanism to point b) but with a single change: after chosing the 3 prefects from the 5 I chose the remaining from the 7 non-prefects AND the 2 remaining prefects:
9 nCr 5 = 126
126 * 10 = 1260
The books says that answer is 420:mad::mad::mad:

What the fuck am I doing wrong?

Thanks!

[LiquidIce]

Fail :facepalm:

The right way to solve this is to consider all the possible cases:
if the committee is to be made up of at least 3 prefects then there are the following cases to consider:
3 prefects + 5 boys
4 prefects + 4 boys
5 prefects + 3 boys

This is to be solved by:
5 nCr 3 * 7 nCr 5 + 5 nCr 4 * 7 nCr 4 + 5 nCr 5 * 7 nCr 3 = 420

I still have to figure out WHY this is so, since my way seems oh so logical, but this, according to the book and a friend, is right.

Case closed (but if any one wants to points out the error in my reasoning, please do so).

[Mantikore]

ahh yes, permutations and combinations, my old arch enemy. its been a while since i tackled these kinds of questions.

youre right about the first one.
you got 12 people, 5 of them are prefects. split the people into 2 groups: prefects and non prefects. this obviously gives 7 non prefects

when you have a committee of 8 that consists of 3 prefects, obviously, the committee has 5 non prefects

7 nCr 5 * 5 nCr 3 = 210, your answer

the second one is a bit more annoying. the methodology in the second post means that there are three sets of combinations: one with 3 prefects (ie the one in part(a)), one with 4 prefects, and one with 5 prefects. just work out the number of combinations of the three, then add them to get the total number of combinations for those three sets.


I apply the exact same solving mechanism to point b) but with a single change: after chosing the 3 prefects from the 5 I chose the remaining from the 7 non-prefects AND the 2 remaining prefects:
9 nCr 5 = 126
126 * 10 = 1260
The books says that answer is 420:mad::mad::mad:


i see what youre getting at with that logic

im not exactly sure, but i think that if you just assume the remaining 2 prefects not selected as part of the 'prefect group' can be considered part of the remaining boys, you would assume that those 2 would always be a part of the remaining boys. thats not always the case, since it could be 1 prefect with the remaining boys, or 0 prefects. by using (7+2) nCr 5, those 2 will always be considered as part of the non prefects group, which gives a high number of combinations on that part

lol i dunno :confused:

[LiquidIce]

My second post gives the right answer but it is still mysterious to me. I tried this method to some other examples and they all worked. The method works with all "at least something" examples.

Now, off to probability, discrete random variables and such.

[harry_hardcore_hoedown]

Fail :facepalm:

The right way to solve this is to consider all the possible cases:
if the committee is to be made up of at least 3 prefects then there are the following cases to consider:
3 prefects + 5 boys
4 prefects + 4 boys
5 prefects + 3 boys

This is to be solved by:
5 nCr 3 * 7 nCr 5 + 5 nCr 4 * 7 nCr 4 + 5 nCr 5 * 7 nCr 3 = 420

I still have to figure out WHY this is so, since my way seems oh so logical, but this, according to the book and a friend, is right.

Case closed (but if any one wants to points out the error in my reasoning, please do so).

LiquidIce is right. I think there's a simpler way, but I can't think of it. The reason is that in your case, imagine the five prefects are A, B, C, D, and E. And when three are being chosen, you could have, for example:

A, B, and C
or A, B and D

If you selected A, B, and C, the other five would be chosen from D, E, x, x, x, x, x, x, x, x. So, you could end up with:

A, B, C, D, x, x, x, x.

If you selected A, B and D, the other five are chosen from C, E, x, x, x, x, x, x, x. You could end up with:

A, B, C, D, x, x, x, x.

They're the same combination, however they're counted as two by your method.