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How many possible 6-digit code numbers can be formed from three a, two

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How many possible 6-digit code numbers can be formed from three a, two [#permalink]

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New post 19 Mar 2017, 18:09
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How many possible 6-digit code numbers can be formed from three a, two b, and one c?

A. 40
B. 50
C. 60
D. 70
E. 80

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Re: How many possible 6-digit code numbers can be formed from three a, two [#permalink]

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New post 19 Mar 2017, 20:46
Total Number of 6-digit code numbers = 6 ! / (3! * 2!) = 60

Answer is C. 60
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Re: How many possible 6-digit code numbers can be formed from three a, two [#permalink]

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New post 19 Mar 2017, 20:54
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MathRevolution wrote:
How many possible 6-digit code numbers can be formed from three a, two b, and one c?

A. 40
B. 50
C. 60
D. 70
E. 80


----------ASIDE-----------------------
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 P's
So, the total number of possible arrangements = 11!/[(4!)(4!)(2!)]

------NOW ONTO THE QUESTION----------
We want to arrange AAABBC
There are 6 letters in total
There are 3 identical A's
There are 2 identical B's
So, the total number of possible arrangements = 6!/[(3!)(2!)]
= 60

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Re: How many possible 6-digit code numbers can be formed from three a, two [#permalink]

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New post 20 Mar 2017, 00:54
MathRevolution wrote:
How many possible 6-digit code numbers can be formed from three a, two b, and one c?

A. 40
B. 50
C. 60
D. 70
E. 80


No of ways = 6!/( 3! . 2!) = 60

Option C
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Re: How many possible 6-digit code numbers can be formed from three a, two [#permalink]

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New post 20 Mar 2017, 03:42
==> You get a,b,b, c,c,c, which is 6!/(2!)(3!)=60.

Therefore, the answer is C.
Answer: C
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Re: How many possible 6-digit code numbers can be formed from three a, two   [#permalink] 20 Mar 2017, 03:42
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How many possible 6-digit code numbers can be formed from three a, two

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