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

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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Posted from my mobile device

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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