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Updated on: 06 Nov 2012, 03:22
Originally posted by xcusemeplz2009 on 23 Dec 2009, 06:38.
Last edited by Bunuel on 06 Nov 2012, 03:22, edited 1 time in total.
Last edited by Bunuel on 06 Nov 2012, 03:22, edited 1 time in total.
Renamed the topic and edited the question.
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Difficulty:
35%
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Question Stats:
71% (01:38) correct
29%
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based on 803
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Find the sum of all the four digit numbers which are formed by digits 1, 2, 5, 6
A. 933510
B. 93324
C. 65120
D. 8400
A. 933510
B. 93324
C. 65120
D. 8400
23 Dec 2009, 07:16
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xcusemeplz2009
The answer choices make the solution easy:
There are 4!=24 four digit numbers which are formed by digits 1, 2, 5, 6.
Obviously 24/4=6 numbers will end with 1; 6 numbers will end with 2, 6 numbers with 5 and 6 numbers with 6.
6*1+6*2+6*5+6*6=6*14=84, which means that the sum of all these 24 numbers must end by 4, only answer choice with 4 at the end is B.
Answer: B.
But if we were not given such an easy answer choices, the solution would be:
We have 24 numbers of the of the form: 1000a+100b+10c+d, where a, b, c and d can take any value from the set {1, 2, 5, 6} and there will be 6 numbers with same digit (a, b, c, d) at the thousands, hundreds, tens and units digits.
(1000*6a+1000*6b+1000*6c+1000*6d)+(100*6a+100*6b+100*6c+100*6d)+(10*6a+10*6b+10*6c+10*6d)+(6a+6b+6c+6d)=(6a+6b+6c+6d)(1111)=
=6(a+b+c+d)(1111)=6*14*1111=93324
Generally the sum of all the numbers which can be formed by using the n distinct digits, is given by the formula:
(n-1)!*(sum of the digits)*(111…..n times)
