First slide

Combinations

Question

$T h e n u m b e r o f s e v e n d i g i t e d i n t e g e r s w i t h s u m o f t h e d i g i t s e q u a l t o 10 a n d f o r m e d b y u \sin g t h e d i g i t s 1, 2 a n d 3 o n l y i s$

Moderate

A

77
B

35
C

82
D

42

Solution

            The given digits are 1,2,3
            The number of seven digited numbers whose sum of the digits is 10 :
Case (i) Suppose that the digits are 1,1,1,1,1,2,3, this can be arranged among themselves in $\frac{7!}{5!} = 42$ ways
Case (ii) Suppose that the digits are 1,1,1,1,2,2,2, this can be arranged among themselves in   $\frac{7!}{4! 3!} = 35$ ways.
Total number of 7 digited numbers is $42 + 35 = 77$

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