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How many different positive integers having six digits are

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How many different positive integers having six digits are  [#permalink]

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New post 12 Mar 2014, 05:10
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A
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Question Stats:

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How many different positive integers having six digits are there, where exactly on 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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Re: How many different positive integers having six digits are  [#permalink]

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New post 12 Mar 2014, 05:22
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idinuv wrote:
How many different positive integers having six digits are there, where exactly on 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


344578 --> permutations of these 6 digits where 2 of them are identical (two 4's) is 6!/2! = 360;

344577 --> permutations of these 6 digits where 2 of them are identical (two 4's) and 2 others are also identical (two 7's) is 6!/(2!2!) = 180;

344588 --> permutations of these 6 digits where 2 of them are identical (two 4's) and 2 others are also identical (two 8's) is 6!/(2!2!) = 180.

Total = 360 + 180 + 180 = 720.

Answer: B.

Hope it's clear.
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Re: How many different positive integers having six digits are  [#permalink]

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New post 09 Dec 2015, 03:29
Bunuel can you please explain in deep ?
I am not getting this :cry:
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Re: How many different positive integers having six digits are  [#permalink]

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New post 09 Dec 2015, 04:14
suresh8055 wrote:
Bunuel can you please explain in deep ?
I am not getting this :cry:


Hey Suresh!

As per my understanding, the concept is about simple arrangement.
Since the question states that each of the last digits are a 7 or an 8 and that there are only two places left at the end.
This gets us to 3 cases: Case 1 - a 7 and then an 8 in the end
Case 2 - a 7 and a 7 in the end
Case 3 -an 8 and an 8 in the end

Now, for each of these cases a random selection of the numbers have been taken because they have to be arranged anyways.
So for Case 1 - We have 344578. As per the permutations concept, since there are repeating items that have to be permuted we divide the number by the frequency of the repeating item.
So it results as 6!/2! (4 repeats twice)

For Case 2 - We have 344577. As explained for Case 1, we have 6!/(2!*2!). Here both 4 and 7 repeat twice.

Similarly, in Case we have 344588. A permutation results 6!/(2!*2!).

The grand total on all the 3 cases is 6!/2!+ 6!/(2!*2!)+6!/(2!*2!) = 360+180+180 = 720

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Re: How many different positive integers having six digits are  [#permalink]

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New post 09 Dec 2015, 22:25
Hi All,

For this question, 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

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Re: How many different positive integers having six digits are  [#permalink]

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New post 08 Feb 2017, 19:47
idinuv wrote:
How many different positive integers having six digits are there, where exactly on 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



We need to determine the number of 6-digit numbers in which 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.

Option #1: The number has two 7s

3-4-4-5-7-7

We use the indistinguishable permutations formula to determine the number of ways to arrange 3-4-4-5-7-7:

6!/(2! x 2!) = (6 x 5 x 4 x 3 x 2)/(2 x 2) = 6 x 5 x 3 x 2 = 180 ways

Option #2: The number has two 8s

3-4-4-5-8-8

6!/(2! x 2!) = 180 ways

Option #3: The number has a 7 and an 8

3-4-4-5-7-8

6!/2! = (6 x 5 x 4 x 3 x 2)/2 = 6 x 5 x 4 x 3 = 360 ways

Thus, the number can be created in 180 + 180 + 360 = 720 ways.

Answer: B
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Re: How many different positive integers having six digits are  [#permalink]

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Re: How many different positive integers having six digits are &nbs [#permalink] 24 Mar 2018, 06:43
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