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permutations problem

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permutations problem

by Malolo355 » Tue Apr 29, 2014 10:51 am
Hi,

I'm having trouble figuring out this question:

How many different positive integers having six digits are there, where exactly one of the digits is a 3,exactly two of the digits are a 4, exactly one of the digits is a 5, and each of the other digits is a 7 or an 8?

A)360
B)720
C)840
D)1,080
E)1,440
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by [email protected] » Tue Apr 29, 2014 11:06 am
Hi Malolo355,

For this questions, there are a couple of different ways to "do the math." Here's a way that breaks down the possibilities into small groups:

Based on the prompt, we know we have:
3, 4, 4, 5, (7 or 8), (7 or 8)

This breaks down into 3 possibilities:
3, 4, 4, 5, 7, 7
3, 4, 4, 5, 7, 8
3, 4, 4, 5, 8, 8

We can now calculate how many 6-digit numbers there are for each possibility.

If we had 6 DIFFERENT numbers (e.g. 1, 2, 3, 4, 5, 6), there would be 6! = 720 different 6-digit numbers

For 3, 4, 4, 5, 7, 7 though, we have some "duplicate numbers" (two 4s and two 7s), which affect the math. Each of those "sets of 2" means that we have to divide by 2!

So, here we'd have 6!/{2!2!] = 720/4 = 180 different 6-digit numbers

For 3, 4, 4, 5, 8, 8 we have the same "duplicate numbers" situation (two 4s and two 8s)....

So, here we'd also have 6!/[2!2!] = 180 different 6-digit numbers

For 3, 4, 4, 5, 7, 8 we have just one group of duplicates (two 4s), so we divide by just 2!....

Here, we'd have 6!/2! = 720/2 = 360 different 6-digit numbers

In total, we have 180 + 180 + 360 = 720 different 6-digit numbers

Final Answer: B

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by Malolo355 » Wed Apr 30, 2014 12:34 pm
That solution seems a little complicated to me and under the test pressure it might be time consuming.
Is there any other solution?
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by GMATGuruNY » Wed Apr 30, 2014 2:07 pm
Malolo355 wrote:Hi,

I'm having trouble figuring out this question:

How many different positive integers having six digits are there, where exactly one of the digits is a 3,exactly two of the digits are a 4, exactly one of the digits is a 5, and each of the other digits is a 7 or an 8?

A)360
B)720
C)840
D)1,080
E)1,440
There are a total of 6 positions in the integer.

Two 4's:
The two 4's must occupy 2 of the 6 positions in the integer.
Number of ways to choose 2 positions from 6 options = 6C2 = (6*5)/(2*1) = 15.

At this point, 4 positions in the integer remain.

3 and 5:
Number of options for the 3 = 4. (Any of the 4 remaining positions.)
Number of options for the 5 = 3. (Any of the 3 remaining positions.)

At this point, 2 positions in the integer remain.

Remaining positions:
Number of options for the first remaining position = 2. (7 or 8.)
Number of options for the last remaining position = 2. (7 or 8.)

To combine the options above, we multiply:
15*4*3*2*2 = 720.

The correct answer is B.
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