The number of ways of distributing 15 identical balloons, 6 identical pencils and 3 identical erasers among 3 children, such that each child gets at least four balloons and one pencil, is

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Event 1: Distribution of balloons

Since each child gets at least 4 balloons, we will initially allocate these 4 balloons to each of them.

So we are left with $15 - 4 \times 3 = 15 - 12 = 3$ balloons and 3 children.

Now we need to distribute 3 identical balloons to 3 children.

This can be done in ${}^{n+r-1}C_{r-1}$ ways, where $n = 3$ and $r = 3$.

So, number of ways = ${}^{3+3-1}C_{3-1} = {}^5C_2 = \frac{5 \times 4}{2 \times 1} = 10$

Event 2: Distribution of pencils

Since each child gets at least one pencil, we will allocate 1 pencil to each child. We are now left with $6 - 3 = 3$ pencils.

We now need to distribute 3 identical pencils to 3 children.

This can be done in ${}^{n+r-1}C_{r-1}$ ways, where $n = 3$ and $r = 3$.

So, number of ways = ${}^{3+3-1}C_{3-1} = {}^5C_2 = \frac{5 \times 4}{2 \times 1} = 10$

Event 3: Distribution of erasers

We need to distribute 3 identical erasers to 3 children.

This can be done in ${}^{n+r-1}C_{r-1}$ ways, where $n = 3$ and $r = 3$.

So, number of ways = ${}^{3+3-1}C_{3-1} = {}^5C_2 = \frac{5 \times 4}{2 \times 1} = 10$

Applying the product rule, we get the total number of ways = $10 \times 10 \times 10 = 1000$.

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