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)