If \[16\] identical pencils are distributed among 4 children such that each gets at least 3 pencils. Then, find the number of ways of distributing the pencils.
A. \[15\]
B. \[25\]
C. \[35\]
D. \[40\]

Hint: First, calculate the number of pencils equally distributed to each child. Then, calculate the number of ways of distributing the remaining identical pencils among the 4 children and get the required answer.

Formula Used: The number of ways of distributing \[n\] identical objects among the \[r\] groups is: \[{}^{n + r - 1}{C_{r - 1}} = \dfrac{{\left( {n + r - 1} \right)!}}{{n!\left( {r - 1} \right)!}}\]

Complete step by step solution: Given:
\[16\] identical pencils are distributed among 4 children
each child gets at least 3 pencils

Here, pencils are identical. So, it doesn’t matter which pencil a child receives.
It is given that each child must receive at least 3 pencils.
So, distribute 3 pencils to each child.
Number of distributed pencils: \[3 \times 4 = 12\]

The remaining number of pencils are: \[16 - 12 = 4\]
Now we must distribute 4 identical pencils to 4 children.
So, apply the formula of the number of ways of distributing \[n\] identical objects among the \[r\] people.
We get,
\[{}^{4 + 4 - 1}{C_{4 - 1}} = \dfrac{{7!}}{{4!\left( {4 - 1} \right)!}}\]
\[ \Rightarrow {}^{4 + 4 - 1}{C_{4 - 1}} = \dfrac{{7 \times 6 \times 5 \times 4!}}{{4! \times 3!}}\]
\[ \Rightarrow {}^{4 + 4 - 1}{C_{4 - 1}} = 35\]
Therefore, the number of ways of distributing \[16\] identical pencils among 4 children such that each child gets at least 3 pencils are \[35\].

Option ‘C’ is correct

Note: The factorial of a number is a product of all whole numbers less than that number up to 1.
Formulas:
\[n! = n\left( {n - 1} \right)\left( {n - 2} \right)....3 \times 2 \times 1\]
\[n! = n\left( {n - 1} \right)!\]