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by vinay1983 » Sun Sep 29, 2013 11:35 pm
How many different 9 letter sequences can be formed such that they contain 2 Ps, 3Bs and 4Cs?

1. 1300
2. 1320
3. 1260
4. 1340
5. 1420
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by faraz_jeddah » Sun Sep 29, 2013 11:39 pm
vinay1983 wrote:How many different 9 letter sequences can be formed such that they contain 2 Ps, 3Bs and 4Cs?

1. 1300
2. 1320
3. 1260
4. 1340
5. 1420
Combinations formula with repeated letters = Total! / eachletter!

= 9! / 4!x3!x2!

= 1260

Answer should be 3
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by rakeshd347 » Sun Sep 29, 2013 11:39 pm
vinay1983 wrote:How many different 9 letter sequences can be formed such that they contain 2 Ps, 3Bs and 4Cs?

1. 1300
2. 1320
3. 1260
4. 1340
5. 1420
There will be total 9! ways to arrange 9 items but because we have 2ps, 3Bs and 4 Cs so we have to divide the 9! by 2!, 3! and 4!.

So answer should be 9!/2!3!4!=1260
OA should be C
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by ganeshrkamath » Mon Sep 30, 2013 12:42 am
vinay1983 wrote:How many different 9 letter sequences can be formed such that they contain 2 Ps, 3Bs and 4Cs?

1. 1300
2. 1320
3. 1260
4. 1340
5. 1420
2+3+4 = 9
Total number of combinations = 9!/(2!3!4!) = (9*8*7*6*5)/(2*2*3)
= 9*4*7*5 = 1260
Choose 3

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by GMATGuruNY » Mon Sep 30, 2013 4:38 am
vinay1983 wrote:How many different 9 letter sequences can be formed such that they contain 2 Ps, 3Bs and 4Cs?

1. 1300
2. 1320
3. 1260
4. 1340
5. 1420
The number of ways to arrange 9 DISTINCT elements = 9!.

But the arrangement here includes IDENTICAL elements: 2 P's, 3 B's, and 4 C's.
When an arrangement includes IDENTICAL elements, we must divide by the number of ways EACH SET of identical elements can be arranged.
The reason:
When the identical elements swap positions, the arrangement doesn't change.
Here, we must divide by 2! (to account for the 2 P's), by 3! (to account for the 3 P's), and by 4! (to account for the 4 C's):
9!/(2!3!4!) = 1260.

The correct answer is C.
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by Brent@GMATPrepNow » Mon Sep 30, 2013 6:36 am
In general, when we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:

If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....]

So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows:
There are 11 letters in total
There are 4 identical I's
There are 4 identical S's
There are 2 identical O's
So, the total number of possible arrangements = 11!/[(4!)(4!)(2!)]

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