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natural number

by shashank.ism » Sat Feb 20, 2010 1:33 pm
How many six-digit numbers can be made by using the first 6 natural numbers such that the digit at the unit's place is greater than the digit at the hundred's place and the number thus formed is a multiple of 4? (Repetition of digits is not allowed.)

A) 96
B) 108
C) 114
D) 78
E) 120
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by lalmanistl » Sat Feb 20, 2010 4:54 pm
My choice is D.

What is the answer?

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by lalmanistl » Sat Feb 20, 2010 5:03 pm
I reevaluated my answer and it comes to 114 now.

What is the right answer? Also i used various combination to find the answer, didn't reach to find a single formula.

Can someone explain the answer too...
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by shashank.ism » Sun Feb 21, 2010 12:35 pm
lalmanistl wrote:I reevaluated my answer and it comes to 114 now.

What is the right answer? Also i used various combination to find the answer, didn't reach to find a single formula.

Can someone explain the answer too...
You are correct lalamnist ....The OA is 114 i.e. AnsC
OS is this
(1, 3, 2), (1, 5, 2), (1, 2, 4), (1, 6, 4), (1, 3, 6), (1, 5, 6)
(2, 6, 4), (2, 1, 6), (2, 3, 6), (2, 5, 6)
(3, 2, 4), (3, 6, 4), (3, 1, 6), (3, 5, 6)
(4, 1, 6), (4, 3, 6), (4, 5, 6)
(5, 1, 6), (5, 3, 6)
⇒ 19 combination
The remaining 3 places in 3! ways i.e. 6 ways
So total number of ways = 6 × 19 = 114 ways
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