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# Combination and maths...............

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Manager
Joined: 02 Sep 2008
Posts: 103

Kudos [?]: 104 [0], given: 1

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27 Apr 2009, 19:28
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Difficulty:

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Question Stats:

100% (01:09) correct 0% (00:00) wrong based on 0 sessions

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Kudos [?]: 104 [0], given: 1

Senior Manager
Joined: 08 Jan 2009
Posts: 325

Kudos [?]: 176 [1], given: 5

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27 Apr 2009, 22:15
1
KUDOS
From the combination u can notice that All the diigits will occur equal no of times in the Units, tens, hundreds positon.

To find how many times they occur equally = No of combinations/ No of digits = 24 / 4 = 6

For the units position alone : 6 ( 1 +2+3+ 4 ) = 60

Thousand + Hundreds + Tens + units = 60 *1000 + 60 * 100 + 60 *10 + 60 = 66660

Kudos [?]: 176 [1], given: 5

Manager
Joined: 02 Mar 2009
Posts: 134

Kudos [?]: 55 [0], given: 0

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27 Apr 2009, 22:47
Excellent tkarthi! Thanks!

I managed to find the answer but by finding the last 3 digits, that is 660. Then picked E. Here's how:

If first two spots are 12 ---combinations for last two 34 + 43 = 77. Multiply by 2 because the combination when 21 one are the first two digits will be the same. Therefore, 77*2= 154
If first two spots are 13 ---combinations for last two 24 + 42 = 66. 66*2 = 132
If first two spots are 14 ---combinations for last two 32 + 23 = 55. 55*2 = 110

If first two spots are 23 ---combinations for last two 14 + 41 = 55. 55*2 = 110
If first two spots are 24 ---combinations for last two 13 + 31 = 44. 44*2 = 88

If first two spots are 34 ---combinations for last two 12 + 21 = 33. 33*2 = 66

154+132+110+110+88+66 = 660

Thus E.

I acknowledge that tkarthi's way is more comprehensive.

Kudos [?]: 55 [0], given: 0

Re: Combination and maths...............   [#permalink] 27 Apr 2009, 22:47
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