combination

Question

The number of seven digit integers possible with sum of the digits equal to 11 and formed by using the digits 1, 2 and 3 only are

Comments

the answer will be,

$[x^{11}]$   in    $(x+x^2+x^3)^7$

$[x^{11}]$   in    $x^7(1-x^3)^7(1-x)^{-7}$

$[x^4]$     in      $(1-x^3)^7(1-x)^{-7}$

$=\binom{-7}{4}+\binom{7}{1}\binom{-7}{1}$

$=210 - 49 = 161$

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Case 1 : 4 one's , 2 two's, 1 three $= \frac{7!}{4!*2!} = 105$

Case 2 : 5 one's, 2 three's $= \frac{7!}{5! * 2!} = 21$

Case 3 : 3 one's, 4 two's $= \frac{7!}{4!*3!} = 35$

Total $105 + 21 + 35 = 161$