The Student Room Group

Combinations problem

Here is the problem:

"Winston must choose 4 classes for his final semester of school. He must take at least 1 science class and at least 1 arts class. If his school offers 4 (distinct) science classes, 3 (distinct) arts classes and 3 other (distinct) classes, how many different choices for classes does he have?"

I'm wondering what's wrong with the following method:

(4x3) choices of first (compulsory) art class and science classes

x

8c2 choices of remaining courses

total 336 (correct answer is 160)?

Thanks
Reply 1
Original post by jameshyland29
Here is the problem:

"Winston must choose 4 classes for his final semester of school. He must take at least 1 science class and at least 1 arts class. If his school offers 4 (distinct) science classes, 3 (distinct) arts classes and 3 other (distinct) classes, how many different choices for classes does he have?"

I'm wondering what's wrong with the following method:

(4x3) choices of first (compulsory) art class and science classes

x

8c2 choices of remaining courses

total 336 (correct answer is 160)?

Thanks


You'd be over/double counting (repetitions) the cases where there was more than one art or science class.

Edit: a slightly long winded way would be to list the combinations of the different classes and then add them up.

S A 0
1 1 2
1 2 1
1 3 0
2 1 1
2 2 0
3 1 0
(edited 5 years ago)
Original post by jameshyland29

Thanks


A reasonably efficient method would be to use inclusion/exclusion.

How many possible choices with no restrictions?

Subtract the number of choices with "no science".

Subtract the number of choices with "no art".

Then add the number of choices with "no art and no science". (There are actually none in this latter category.)
(edited 5 years ago)
Reply 3
Original post by ghostwalker
A reasonably efficient method would be to use inclusion/exclusion.

How many possible choices with no restrictions?

Subtract the number of choices with "no science".

Subtract the number of choices with "no art".

Then add the number of choices with "no art and no science". (There are actually none in this latter category.)

Certainly more elegant than my long winded way.
Thanks for that I've just about got to grips with the PIE now. :u:
Reply 5
Original post by jameshyland29
Thanks for that I've just about got to grips with the PIE now. :u:

Good, its a useful thing.
A "sanity check" I noticed at the time was that 10C4 is 210 which would be larger than your original answer which includes extra restrictions. Double (or higher) counting is very easy to do, so its always worth doing a sanity check when you don't know the answer.

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