Wednesday 9 November 2011

P&C-10 (IIT-JEE 2009)

The number of seven digit integers, with sum of the digits equal to 10 and formed by using the digits 1,2, and 3 only is
(a)55          (b) 66              (c)77                    (d)88
Solution follows here:
Solution:
Only possibilities:
1's
2's
3's
Number of digits
Sum of digits
5
1
1
7
10
4
3
0
7
10
Permutations for the first case (five 1’s, one 2 and one 3) = 7!/5! = 7*6 = 42
Permutations for the second case (four 1’s, three 2’s) = 7!/(4!3!) = (7*6*5)/(3*2) = 35
Total number of possibilities = 42+35 = 77
Answer(c)

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