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Permutations and Combinations general question

Problem Solving — algebra and arithmetic
Expert replies
by Osirus@VeritasPrep » Tue Oct 20, 2009 7:23 am
Can someone explain when you have a P&C problem that has multiple sets within the problem when do you multiply the sets together to get the answer and when do you add the sets to get the answer?

Examples:

1) The Natural Woman, a women's health food store, offers its own blends of trail mix. If the store uses 4 different ingredients, how many bins will it need to hold every possible blend, assuming that each blend must have at least two ingredients? (Also, assume that each bin can hold one and only one blend)

2) A pod of 6 dolphins always swims single file, with 3 females at the front and 3 males in the rear. In how many different arrangements can the dolphins swim?

With the first question you add 4!/(2!2!) + 4!/3! + 4!/4!

With the second question you multiply 3! * 3!

In general, how can you tell when you would add vs when you would multiply the sets?

Thanks in advance.
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Source: — Quantitative Reasoning |

by GMATBootcamp » Tue Oct 20, 2009 11:52 am
Hi osirus0830,


You add when you break the outcome into different cases. You add the cases to get the total.

You multiply when you take a sequence of actions to get the desired outcome.


In your first example, you are breaking the scenario down into three cases.

case1 - the # of bins required to house 2 ingredient blends
case2 - the # of bins required to house 3 ingredient blends
case3 - the # of bins required to house 4 ingredient blends.

We add all three cases together to find the total # of bins required

The second example follows a sequence of actions: first we choose female dolphins, then we choose male dolphins.

There are 3! possible combinations for female dolphins and 3! possible combinations for male dolphins. We multiply them together to find the total number of possible combinations.


HTH!
Paul

Focused GMAT Preparation
www.thegmatbootcamp.com
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by Osirus@VeritasPrep » Tue Oct 20, 2009 12:05 pm
Thanks, I appreciate the reply
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Hey

by Leon1984 » Wed Oct 21, 2009 3:36 pm
For me, it helps to imagine, talk to myself.

With the first, I got locked on the word "at least".
So it was like this: Choose 2 out of 4, but wait, I can also choose 3 out of 4, and also 4 out of 4. So calculate the number of ways to make each choice, and sum.

In the second, we are not choosing, just arranging. So, how can I arrange the first 3? How can I arrange the other 3? And since we are arranging, for every one arrangement of the first 3 I can have all the numbers of arrangement of the other 3, so we multiply to get the total way to arrange 2 groups.

Hope it helps...
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by Mayab » Mon Jun 17, 2013 3:21 am
I am trying to solve the first question by using the following method; however it is not giving me the same reply.
Since this is an " at least' question: the total number of arrangements less the undesired outcomes should give us the desired outcome. IN this case, the total number of arrangements should be 4!=4*3*2*1=24 and the undesired outcome is to have 1 ingredient and zero ingredient. Having 1 ingredient is equivalent to 4!/3!=4 and having zero ingredients is equivalent to 4!/4!=1 ==> the total undesired outcomes=4+1=5 less the total arrangements =24, which results in the number of bins=24-5=19 . However the answer that you are giving is 11. Where did I go wrong in my analysis?? Please advise. Thanks!
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