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tough sequences

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by dmateer25 » Wed Dec 10, 2008 3:00 pm
6!/2!3! = 60
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by albema » Wed Dec 10, 2008 3:07 pm
dmateer25 wrote:6!/2!3! = 60
Can you explain this, please?
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by dmateer25 » Wed Dec 10, 2008 3:36 pm
6 letters can be arranged 6! ways

However, there are 2 B's and 3 C's.

In order to eliminate duplicate arrangements, you must divide by 3! (for the 3 C's)* 2! (For the 2 B's)
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Re: tough sequences

by canuckclint » Wed Dec 10, 2008 6:44 pm
albema wrote:How many different 6-letter sequences are there that consist of 1 A, 2 B’s, and 3 C’s ?

A. 6
B. 60
C. 120
D. 360
E. 720


B
Yes the above is the standard approach. Another intuitive approach is
to have the alphabets into 6 slots.

A BB CCC

For A its 6 c 1 = 6, 6 slots and one A to choose from.
For B its 5 c 2 = 10, 5 slots that can be filled with 2 Bs
For C its 3 c 3 = 1, 3 slots left and 3 Cs to fill with


6 * 10 * 1 = 60

--
Clinton

Bsc Computer Science
Math Minor
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by jnellaz » Thu Dec 11, 2008 8:06 am
First try to find ALL the combinations a 6 letter sequence. Which is equal to: 6 * 5 *4 * 3 * 2 * 1 or 6!.

Then we can try to eliminate duplicates in the sequence. If we break the sequence of letters as follows:

1 2 3 4 5 6
A B1 B2 C1 C2 C3

We can see that there are cases where there will be duplicate sequences. So as a quick example:

A B(1) B(2) C1 C2 C3 will be the same as
A B(2) B(1) C1 C2 C3.

So we have 2 B's = 2! combinations
And 3 C's = 3! combinations

If we divide our original total combinations of a six letter sequence: 6! by 2! and 3! we will get a total of 60 combinations.

Hence - B.
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