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Permutation question

by sk02 » Sun Nov 08, 2009 9:59 pm
Hi, could someone provide a more detailed answer for the following permutation question:
Q. How many six-letter sequences consisting of 1A, 2Bs and 3Cs are possible?
The answer guide gave the following:
If all letters in sequence different, 6! sequences possible.
But, arrangement of 2Bs and 3Cs need to be removed, therefore ans = 6! /((2!)(3!))= 60.
I'm having trouble understanding why, to remove the arrangement of 2Bs and 3Cs, the division needs to occur. Thanks. [/url]
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by palvarez » Mon Nov 09, 2009 7:30 am
Treat 2B's and 3C's differently.

B1, B2
C1, C2, C3.

In this case, you have 6! cases.

However, note that B1B2 = B2B1, C1C2 = C2C1, C1C2C3 = C3C2C1, etc.

So, you need to get rid of these redundant entries.

That's why you gotta divide 6! by 2! (for B1 and B2), and by 3! (for C1, C2 and C3, since they are indistinguishable).
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Re: Permutation question

by Brent@GMATPrepNow » Mon Nov 09, 2009 2:52 pm
sk02 wrote:Hi, could someone provide a more detailed answer for the following permutation question:
Q. How many six-letter sequences consisting of 1A, 2Bs and 3Cs are possible?
The answer guide gave the following:
If all letters in sequence different, 6! sequences possible.
But, arrangement of 2Bs and 3Cs need to be removed, therefore ans = 6! /((2!)(3!))= 60.
I'm having trouble understanding why, to remove the arrangement of 2Bs and 3Cs, the division needs to occur. Thanks. [/url]
In general, the number of permutations of n objects, when a are alike, another b are alike, another c are alike and so on, is:
n!/(a!b!c!...)
Brent Hanneson - Creator of GMATPrepNow.com
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