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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 ________.
in Combinatory by (239 points)
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161...?
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any efficient way to solve this question ??

it's really very time consuming :|
+1
there are 3 groups only
1)(3,3,1,1,1,1,1)
2)(2,2,3,1,1,1,1)
3)(2,2,2,2,1,1,1)
so total number of integers-

$\frac{7!}{2!5!}+\frac{7!}{2!4!}+\frac{7!}{4!3!} = 161$
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@MiniPanda @Shubhgupta

Yes, the answer given is 161. The video solution given, talked about some generating function problem. Know any thing about it?
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Yeah I did using generating solution..

Though solution of @Shubhgupta

 is easier but it might take time to think of all combinations..

i will post the solution after some time

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@Magma @VikramRB I added the answer..let me know if anything is not understandable.

1 Answer

+1 vote

$x_1,x_2,x_3,x_4,x_6,x_7$ be 7 digits each of whom can take 1 or 2 or 3. We need to collect 11 from their sum.
Let me represent value=1 as c, value=2 as $c^2$ , value= 3 as $c^3$.

$(c^1 + c^2 + c^3)$ shows all the possible contribution of one digit.

$(c^1+c^2+c^3)(c^1+c^2+c^3)(c^1+c^2+c^3)(c^1+c^2+c^3)(c^1+c^2+c^3)(c^1+c^2+c^3)(c^1+c^2+c^3)$

These are the contribution of 7 digits and all we need to do is to find only those combinations which can make degree of c=11

This would give us the ways in which we can collect 11.
$(c^1+c^2+c^3)^7 = (c(c^0+c^1+c^2))^7$
$=c^7 (c^0+c^1+c^2)^7 ---->$ from here we need to get coefficient of $ [c^{11}]$
$=(1+c^1+c^2)^7 ----> [c^4]$

$=( (1+c^1)+c^2)^7 ----> [c^4]$

In how many ways can we get $c^4$ from here:

1) $\binom{7}{2} (1+c^1)^5 (c^2)^2$ 

$(1+c^1)^5 = \binom{5}{0} (c^1)^0$

So, $\binom{7}{2} * \binom{5}{0}=21$ is the coefficient of $c^4$

2)

$\binom{7}{1} (1+c^1)^6 (c^2)^1$ 

$(1+c^1)^6 = \binom{6}{2} (c^1)^2$

So, $\binom{7}{1} * \binom{6}{2} = 105$ is the coefficient of $c^4$

3)

$\binom{7}{0} (1+c^1)^7 (c^2)^0$ 

$(1+c^1)^7 = \binom{7}{4}(c^1)^4$

So, $\binom{7}{0} * \binom{7}{4}=35$ is the coefficient of $c^4$

Total sum is 21+105+35=161

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