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by gmat6087 » Sat Oct 20, 2012 12:55 am
If each permutation of digits 1,2,3,4,5,6 is listed in increasing order(only from given digits and none is repeated) of magnitude, the 289th term will be?

1)326541
2)341256
3)356241
4)314256

OA 2

**Note this is a ranking problem. will such kind of question be asked on GMAT?
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by Jim@StratusPrep » Sat Oct 20, 2012 6:21 am
Haven't seen anything like this, but I wouldn't put it out of the question. Here is how to go through it.

Number of terms that start with each digit is 120:

1 x 5 x 4 x 3 x 2 x 1 = 120

120 start with 1
120 start with 2

Total is 240.

Numbers that start with 31 = 1 x 1 x 4 x 3 x 2 x 1 = 24
Numbers that start with 32 = 24

Sum thus far is 288 (240 + 24 +24)

The next number with start 34, so the answer is 2.
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by gmat6087 » Sat Oct 20, 2012 10:35 pm
Jim@StratusPrep wrote:Haven't seen anything like this, but I wouldn't put it out of the question. Here is how to go through it.

Number of terms that start with each digit is 120:

1 x 5 x 4 x 3 x 2 x 1 = 120

120 start with 1
120 start with 2

Total is 240.

Numbers that start with 31 = 1 x 1 x 4 x 3 x 2 x 1 = 24
Numbers that start with 32 = 24

Sum thus far is 288 (240 + 24 +24)

The next number with start 34, so the answer is 2.
A very clear and crisp explanation. Thanks for the response. Life of gmattakers becomes so easy because of people such as you.

Regards,
Satya.
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