How many ways can the five digits 3, 3, 4, 5, 6 be arranged

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How many ways can the five digits 3, 3, 4, 5, 6 be arranged into a 5-digit number so that the two occurrences of the digit 3 are separated by at least one other digit?

(A) 48
(B) 36
(C) 24
(D) 18
(E) 12

Please, I need experts to help me out with the solution to this question. Thanks

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by [email protected] » Sun Mar 25, 2018 2:55 pm
Hi Roland2rule,

We're asked for the number of different ways that the five digits 3, 3, 4, 5, 6 be arranged into a 5-digit number so that the two occurrences of the digit 3 are separated by at least one other digit. This question can be solved in a couple of different ways; you might find it helpful to break the total down into 'pieces'

Each of the non-3 digits can be in ANY spot, so there will be (3)(2)(1) = 6 options for every placement of the two 3s. Based on the 'restrictions in the prompt, the two 3s could be in any of the following spots:
1st and 3rd
1st and 4th
1st and 5th
2nd and 4th
2nd and 5th
3rd and 5th

With 6 possible placements for the two 3s and 6 options for each placement, there are (6)(6) total permutations.

Final Answer: B

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by Brent@GMATPrepNow » Sun Mar 25, 2018 3:12 pm
Roland2rule wrote:How many ways can the five digits 3, 3, 4, 5, 6 be arranged into a 5-digit number so that the two occurrences of the digit 3 are separated by at least one other digit?

(A) 48
(B) 36
(C) 24
(D) 18
(E) 12

Please, I need experts to help me out with the solution to this question. Thanks
One approach here is to first IGNORE the restriction (about the 3's not appearing together) and then subtract all of the outcomes that break that restriction. I'm sure someone will show that approach.

Here's another approach:

Take the task of arranging the 5 digits and break it into stages.

Stage 1: Arrange the 4, 5 and 6
We can arrange n unique objects in n! ways
So, we can arrange these 3 digits in 3! ways (6 ways)
So, we can complete stage 1 in 6 ways

TRICKY PART: We'll now add some spaces where the 3's can be placed.
So, for example, if in stage 1, we arranged three digits as 645, then we'll add spaces before and after each digit.
So, we'd get: _6_4_5_
We will place the two 3's in two of the 4 possible spaces.
This will ENSURE that the 3's are not together.

Stage 2: Select two spaces in which to place the 3's
Since the order in which we select the spaces does not matter, we can use combinations.
We can select 2 spaces from 4 spaces in 4C2 ways (6 ways)
So, we can complete stage 2 in 6 ways

By the Fundamental Counting Principle (FCP), we can complete the two stages (and thus arrange all 5 digits) in (6)(6) ways (= 36 ways)

Answer: B
--------------------------

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MEDIUM
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DIFFICULT
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- https://www.beatthegmat.com/combinations-t123249.html


Cheers,
Brent
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by GMATGuruNY » Mon Mar 26, 2018 3:01 am
Roland2rule wrote:How many ways can the five digits 3, 3, 4, 5, 6 be arranged into a 5-digit number so that the two occurrences of the digit 3 are separated by at least one other digit?

(A) 48
(B) 36
(C) 24
(D) 18
(E) 12
Good arrangements = (all possible arrangements) - (bad arrangements).

All possible arrangements:
Number of ways to arrange 5 digits = 5! = 120.
But the five digits here include two identical 3'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 two identical 3's:
120/2! = 60.

Bad arrangements:
In a BAD arrangement, the two 3's are in adjacent positions.
To count the arrangements in which the two 3's are in adjacent positions, place the two 3's in a BLOCK, yielding the following list of 4 elements:
[33], 4, 5, 6
The number of ways to arrange the 4 elements {33}, 4, 5 and 6 = 4! = 24.

Good arrangements:
(all possible arrangements) - (bad arrangements) = 60 - 24 = 36.

The correct answer is B.
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by swerve » Mon Mar 26, 2018 7:22 am
Hi Roland2rule,

Arrangement without restriction = 5! / 2!

Arrangement with the restriction where both the I's are together = 4! 2! / 2! = 4!

(Method: subtracting what's not allowed from total)

Total = (5! / 2!) - 4! = 36. Option B.

Regards!