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How many solutions are there to the inequality x1 + x2 + x3 ≤ 11 where x1, x2, x3 are nonnegative integers? Hint: introduce a variable x4 such that x1 + x2 + x3 + x4 = 11.

Ans. C(4+11-1,11).

My doubt is if the question had been x1 + x2 + x3 >= 11. then how would have we solved it ?
in Combinatory by Loyal (7k points) | 94 views
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@Na what is answer given??

My answer not matching with u
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There would be infinitely many solutions then.

Just take any of the variables to be 11 - the other two become unbounded then.
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@ goxul , can u explain bit that how infinite many solution ?

@ srestha C(4+11-1,11) = C(14,3) = 364
+1
yes Same with generating function too

$\frac{1}{\left ( 1-x \right )^{4}}$

$\frac{4.5.6.7........14}{11!}=364$

Sorry prev comment was wrong
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@srestha mam

@Na462

how to solve it using generating function?

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@himgta

Refer : https://gateoverflow.in/246702/permutation-and-combination

My below comment i solved and discussed the approach using Generating functions

1 Answer

+2 votes
We have to solve $x_1 + x_2 + x_3 \geq 11$, under $\mathbb{N}$.

Notice that if either of the variable is greater than or equal to 11, the other two variables don't matter anymore - the statement has become true. So no matter what value you put to the other two variables, the statement will remain true.

Hence, we have infinite many solutions.
by Loyal (6.8k points)
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