Find the number of 6 digit numbers whose sum is 10 using the digits 1, 2, 3 only

Find the number of 6 digit numbers whose sum is 10 using the digits 1, 2, 3 only

A
60
B
15
C
90
D
100

Solution:

Given digits are 1, 2, 3 sum is 10

Case (i): 1, 1, 1, 1, 3, 3

Number of arrangements $\frac{6!}{4! 2!} = 15$

Case (ii): 1, 1, 2, 2, 2, 2

Number of arrangements $\frac{6!}{4! 2!} = 15$

Code (iii): $1, 1, 1, 3, 2, 2$ $\frac{6!}{4! \times 2!} = \frac{6 \times 5 \times 4!}{4! \times 2} = 15$

$= \frac{6!}{3! \times 2!} = \frac{6 \times 5 \times 4 \times 3!}{3! \times 2 \times 1} = 60$

Total number of arrangements $= 15 + 15 + 60 = 90$

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