JEE Mathematics 2024 Question with Solution

Given: $21$ identical apples are to be distributed among $3$ children, and each child must get at least $2$ apples.

Find: The number of possible distributions.

Use the stars and bars method after satisfying the minimum condition for each child.

First, give $2$ apples to each child.

$$3 \times 2 = 6$$

So the remaining apples are

$$21 - 6 = 15$$

Now let the additional apples received by the three children be $x_1, x_2, x_3$. Then

$$x_1 + x_2 + x_3 = 15$$

where $x_1, x_2, x_3 \ge 0$.

The number of non-negative integer solutions is

$$\binom{15+3-1}{3-1} = \binom{17}{2}$$

Now,

$$\binom{17}{2} = \frac{17 \times 16}{2 \times 1} = \frac{272}{2} = 136$$

Therefore, the number of ways is $136$. The correct option is D.

Given: Each child must receive at least $2$ apples out of $21$ identical apples.

Find: Total valid distributions.

To ensure the condition is satisfied, reserve $2$ apples for each child at the start. This accounts for

$$6$$

apples in total.

The problem is reduced to distributing the remaining

$$15$$

identical apples among $3$ distinct children with no restriction.

For distribution of $n$ identical objects among $r$ distinct groups, the count is

$$\binom{n+r-1}{r-1}$$

Here,

$$n = 15, \qquad r = 3$$

So the required number is

$$\binom{15+2}{2} = \binom{17}{2}$$

Evaluating,

$$\binom{17}{2} = \frac{17 \times 16}{2} = 136$$

Hence, the required number of ways is $136$.