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How to identify perm or comb problem and with/w/o Repetition [#permalink]
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09 May 2010, 01:45
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Is there an easy to way to find out whether the problem is permutation (with or without repetition) or combinataion (with or without repetition)...? any examples would be appreciated.. when is the situation, they mix both perm and comb.....
Apprecaiate your response.



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Re: How to identify perm or comb problem and with/w/o Repetition [#permalink]
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12 May 2010, 17:50
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Hi,
You should use a permutation when the order in which you are suppose to choose a number of objects from a set matters. As an example: In how many ways is it possible to arrange the letters of the word CAT in different 2letter groups, where CA is different than AC (i.e., the order matters)?
\(3P52=\frac{3!}{(32)!}=\frac{3.2.1}{1!}=6\). The general permutation formula is given by \(nPm=\frac{n!}{(nm)!}\). When both elements of the permutation are equal), \(nPn=n!\).
If the order in which the objects are chosen doesn't matter, you use combinations  the formula is very similar to the permutations formula, but you find one more factorial in the denominator: \(nCm=\frac{n!}{(nm)!m!}\). Taking the same example, how many different 2letter groups is it possible to get from the word CAT, considering that CA is the same as AC (i.e., the order doesn't matter)?
\(3C2=\frac{3!}{(32)!2!}=\frac{3.2.1}{1!2.1}=\frac{6}{2}=3\)
If there is repetition, i.e, if the is more than one particular element in the set, you should divide the permutation/combination value by the factorial of the number of objects that are identical. Examples:
How many different 5letter words can be formed from the word APPLE? (note you have 2 Ps).
\(\frac{5P5}{2!}=\frac{5.4.3.2.1}{2.1}=\frac{120}{2}=60\)
How many different 6digit numbers can be written using all of the following six digits: 4,4,5,5,5,7? (here you have 2 fours and 3 fives).
\(\frac{6P6}{2!3!}=\frac{6.5.4.3.2.1}{2.1.3.2.1}=\frac{720}{12}=60\).
Hope this helps!



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Re: How to identify perm or comb problem and with/w/o Repetition [#permalink]
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13 May 2010, 23:11
Thanks for the reply...Can you share any example where perm and comb used together... I understand 'order' plays the role..but from the question, I cant really figure that out.. May be missing something..
thanks again.



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Re: How to identify perm or comb problem and with/w/o Repetition [#permalink]
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13 May 2010, 23:53
Very nice post! +1. pepemelo wrote: Hi,
You should use a permutation when the order in which you are suppose to choose a number of objects from a set matters. As an example: In how many ways is it possible to arrange the letters of the word CAT in different 2letter groups, where CA is different than AC (i.e., the order matters)?
\(3P52=\frac{3!}{(32)!}=\frac{3.2.1}{1!}=6\). The general permutation formula is given by \(nPm=\frac{n!}{(nm)!}\). When both elements of the permutation are equal), \(nPn=n!\).
If the order in which the objects are chosen doesn't matter, you use combinations  the formula is very similar to the permutations formula, but you find one more factorial in the denominator: \(nCm=\frac{n!}{(nm)!m!}\). Taking the same example, how many different 2letter groups is it possible to get from the word CAT, considering that CA is the same as AC (i.e., the order doesn't matter)?
\(3C2=\frac{3!}{(32)!2!}=\frac{3.2.1}{1!2.1}=\frac{6}{2}=3\)
If there is repetition, i.e, if the is more than one particular element in the set, you should divide the permutation/combination value by the factorial of the number of objects that are identical. Examples:
How many different 5letter words can be formed from the word APPLE? (note you have 2 Ps).
\(\frac{5P5}{2!}=\frac{5.4.3.2.1}{2.1}=\frac{120}{2}=60\)
How many different 6digit numbers can be written using all of the following six digits: 4,4,5,5,5,7? (here you have 2 fours and 3 fives).
\(\frac{6P6}{2!3!}=\frac{6.5.4.3.2.1}{2.1.3.2.1}=\frac{720}{12}=60\).
Hope this helps!
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Re: How to identify perm or comb problem and with/w/o Repetition [#permalink]
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21 May 2010, 07:28
pepemelo wrote: \(3P52=\frac{3!}{(32)!}=\frac{3.2.1}{1!}=6\).
Should be: \( 3P2=\frac{3!}{(32)!}=\frac{3.2.1}{1!}=6\). Or am I missing something?



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Re: How to identify perm or comb problem and with/w/o Repetition [#permalink]
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21 May 2010, 07:40
Yes, you are right! It was a typing mistake.
Thanks!



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Re: How to identify perm or comb problem and with/w/o Repetition [#permalink]
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21 May 2010, 07:42
Thank goodness! I was writing out the problem, and I couldn't figure out where the "52" was coming from. Thanks for the explanation, it is terrific!



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Re: How to identify perm or comb problem and with/w/o Repetition [#permalink]
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23 May 2010, 02:12
What would you use with the following problem? (Going off of memory, so I apologize if it sounds weird.)
You have four left socks and four right socks. All of the eight socks are jumbled into one big pile. If you draw socks out at random, what is the probability that you are able to make two pairs of left socks only, and two pairs of right socks only from the entire set of eight socks?
Let me know if I need to be more descriptive and I'll try to locate the problem...




Re: How to identify perm or comb problem and with/w/o Repetition
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