The number of ways of distributing $20$ identical balloons among $4$ children such that each child gets some balloons but no child gets an odd number of balloons, is

We need to distribute 20 identical balloons among 4 children where each child gets some balloons (not zero) and each child gets an even number of balloons.

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Let's say the four children get $a$, $b$, $c$, and $d$ balloons respectively.

Since we're distributing all 20 balloons:

$a + b + c + d = 20$

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If each child gets an even number of balloons, we can write:

Child 1 gets $2x$ balloons (where $x$ is a positive integer)

Child 2 gets $2y$ balloons (where $y$ is a positive integer)

Child 3 gets $2z$ balloons (where $z$ is a positive integer)

Child 4 gets $2p$ balloons (where $p$ is a positive integer)

Since each child must get some balloons, and they must get an even number, the minimum is 2 balloons (when $x = 1$).

$2x + 2y + 2z + 2p = 20$

$x + y + z + p = 10$

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Now we have: find the number of ways to write 10 as a sum of 4 positive integers.

Since $x, y, z, p \geq 1$, let's substitute:

$A = x - 1$ (so $A \geq 0$)

$B = y - 1$ (so $B \geq 0$)

$C = z - 1$ (so $C \geq 0$)

$D = p - 1$ (so $D \geq 0$)

$(A + 1) + (B + 1) + (C + 1) + (D + 1) = 10$

$A + B + C + D = 6$

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We need the number of non-negative integer solutions to:

$A + B + C + D = 6$

For $n$ identical objects distributed among $k$ groups, the number of ways is ${{}}^{{n+k-1}}C_{{k-1}}$.

In our case: $n = 6$, $k = 4$

Number of ways = ${{}}^{{6+4-1}}C_{{4-1}} = {{}}^{{9}}C_{{3}}$

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${{}}^{{9}}C_{{3}} = \dfrac{9!}{3!(9-3)!} = \dfrac{9 \times 8 \times 7}{3 \times 2 \times 1} = \dfrac{504}{6} = 84$

Therefore, there are 84 ways to distribute the balloons.