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Find the number of non negative integral solutions of the equation $$x + y + z = 10$$.

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Hint: Number of non-negative integral solutions of this type of equations $${x_1} + {x_2} + ............ + {x_n} = k$$ can be calculated as number of ways in which k identical balls can be distributed into n distinct boxes will be equal to $$^{(k + n - 1)}{C_{(n - 1)}}$$.

Complete Answer:
We know that
The number of positive integral solutions of the equation $${x_1} + {x_2} + ............ + {x_n} = k$$ is $$^{(k + n - 1)}{C_{(n - 1)}}$$⋅
Here it is given the equation $$x + y + z = 10$$
So the value of k is equal to 10.
And the value of n is equal to 3.
Use the formula $$^{(k + n - 1)}{C_{(n - 1)}}$$
On substituting the value of k and n we get
$$^{(10 + 3 - 1)}{C_{(3 - 1)}}$$
On simplifying the above results
$$^{(12)}{C_{(2)}}$$
On further calculation
$$ = \dfrac{{12!}}{{10! \times 2!}}$$
$$ = \dfrac{{12 \times 11}}{2}$$
$$ = 66$$

Hence the number of non negative positive integral solutions of $$x + y + z = 10$$ is equal to 66.

Note: Here, the idea is when we add a + b + c + ... it should be either all positive or all non negative (which includes zero). You cannot have both in the same equation. What we do at first is, make all the coefficients of variables as whole numbers we can also make it as natural numbers too. Then solve it, but both should give the same result.