A box containing 10 identical compartments has 6 red balls and 2 blue balls. If each compartment can hold only one ball, then the number of different possible arrangements are

Question

Prepp Team · Accepted Answer

Arrangement Calculation: Balls in Compartments

The problem asks for the number of distinct ways to arrange 6 red balls and 2 blue balls into 10 identical compartments, with each compartment holding at most one ball. This implies that $10 - (6 + 2) = 2$ compartments will remain empty.

This is a problem of permutations with repetitions. We have 10 positions (compartments) to fill with 6 identical red balls (R), 2 identical blue balls (B), and 2 identical empty spaces (E).

Permutation Formula

The formula for permutations with repetitions is:

$ \frac{n!}{n_1! n_2! \cdots n_k!} $

Where:

$n$ is the total number of items/positions (here, 10 compartments).
$n_1, n_2, \ldots, n_k$ are the counts of each type of identical item.

Applying the Formula

In this case, $n = 10$. The identical items are:

6 red balls ($n_1 = 6$)
2 blue balls ($n_2 = 2$)
2 empty compartments ($n_3 = 2$)

Plugging these values into the formula:

$ \text{Number of arrangements} = \frac{10!}{6! \times 2! \times 2!} $

Calculation Steps

Calculate the factorials:
- $10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 3,628,800$
- $6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$
- $2! = 2 \times 1 = 2$
Substitute the factorial values into the formula: $ \frac{3,628,800}{720 \times 2 \times 2} $
Simplify the denominator: $720 \times 2 \times 2 = 720 \times 4 = 2880$
Perform the division: $ \frac{3,628,800}{2880} $
Alternatively, simplify before calculating large factorials:
$ \frac{10 \times 9 \times 8 \times 7 \times 6!}{6! \times (2 \times 1) \times (2 \times 1)} = \frac{10 \times 9 \times 8 \times 7}{2 \times 2} = \frac{5040}{4} = 1260 $