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md.sharifulislam01915
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Permutation and combination

by md.sharifulislam01915 Mon Jan 13, 2014 10:57 am

There are 6 books of which 2 are novels.How many ways to select 4 books if 2 of them must be novel?
kwame.appau
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Re: Permutation and combination

by kwame.appau Mon Jan 13, 2014 6:29 pm

-There are 6 ways of selecting 4 books of which 2 are novels.
-This is how to solve it.
-since the question ask for "selection" then it is combination problem.
-Therefore think of 4 slots and lets name the books (A,B,C,D,E,F) and the novels are "A" and "B"
-slots 1 and 2 for "A" and "B" and then slot 3 and 4 for (C,D,E,F)
-Since "A" and "B" are constant, there are 4 ways of selecting any of the remaining books for slot 3 and there are 3 ways of selecting any of the remaining books for slot 4. Hence 4*3/2!= 12/2!. The 2! is the number of ways of arranging books for slot 3 and 4.
-NOTE: order is not important so selecting (A,B,C,D) is the same as selecting (A,B,D,C).
- Hope that helps!
tommywallach
Manhattan Prep Staff
 
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Re: Permutation and combination

by tommywallach Wed Jan 15, 2014 3:36 pm

Hey Guys,

Cool explanation already, but I like to do these ALWAYS with the slots method.

1. Find # of slots (decision points)
2. Fill in each slot with the # of choices at that decision point
3. Determine if order matters or not (or a combination of the two)
4. Multiply the slots together. Divide by the # of slots factorial wherever order doesn't matter.

There aren't actually 4 slots here, because 2 of the slots MUST be filled with the two novels. So there are only 2 slots, because we can only make 2 decisions.

_ _

Now, we have 4 choices for the first slot, and 3 for the second.

4 * 3

However, order doesn't matter, so we divide by the # of slots factorial, or 2!

(4 * 3) / 2! = 6

Hope that helps!

-t