With gaps of one:
1. ( 3 _ 3 _ _ )
2. ( _ 3 _ 3 _ )
3. ( _ _ 3 _ 3 )

With gaps of two:
4. ( 3 _ _ 3 _ )
5. ( _ 3 _ _ 3 )

With gaps of three:
6. (3 _ _ _ 3 )

Total- 6 ways
Remaining numbers: 4, 5, 6- arranged in 3! ways
Therefore 6*3! = 36, Option B

5!/2! - 4! = 120/2 - 24 = 36
B

Option B - 36



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

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The only way that the two digits are separated by at least one other digit is if they are NOT next to each other. We can use the formula:

(Total number of ways to create the numbers) - (number of ways with the 3’s together) = number of ways to create the numbers with 3’s separated by at least one digit

Using the indistinguishable permutations formula, we note that the two 3’s are indistinguishable. Thus, the total number of ways to create the 5-digit numbers is 5!/2! = 60 ways.

Total number of ways to create the numbers with the 3’s together is 4! = 24 ways.

So, the number of ways to create the numbers with the 3’s separated by at least one digit is 60 - 24 = 36.

Answer: B
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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

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

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Hi Brent,

Can this be done using restrictions method too ?

5 numbers with 2 3's can be arranged in 5!/2! ways(using MISSISSIPPI rule) = 60
Restriction: allowing 2 3's to be together, the 4 entities can then be arranged in 4! ways = 24

---> number of ways 2 3's will be separated by atleast one other digit : 60-24 = 36

Thoughts ?

Thanks,
K

That's a perfectly valid (and quick!) solution, Karaan.
Nice work!

Cheers,
Brent
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Since the digit 3 is repeated once,
Total number of 5-digit numbers possible = \(\frac{5! }{ 2!}\) = \(120 / 2\) = 60.

Of these 60 numbers, there will be numbers where the 3s are together. When some objects are to be considered together, we always take them as one object. If we take the 3s as one object, we have 4 objects in total.
Total number of 5-digit numbers where the 3s are together = 4! = 24.

Therefore, total number of 5-digit numbers where the 3s are separated by at least one digit = 60 – 24 = 36.

The correct answer option is B.

Hope that helps!
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Total ways to arrange all given numbers: \(\frac{5! }{ 2!} = \frac{120 }{ 2} = 60\)

=> 3's are together: 4, 5, 6, (3,3) = \(\frac{4! * 2! }{ 2!} = \frac{24 }{ 2} = 24\)

Number of ways in which 3 are separated: 60 - 24 = 36

Answer B
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