Question

25 distinct objects are divided into 5 groups and each group consists of exactly 5 objects. Then the number of ways of forming such groups, is

• A. $\frac{25!}{(5!)^5}$
• B. $\frac{25!}{5!}$
• C. $\frac{25!}{(5!)^6}$
• D. $\frac{25!}{(5!)^4}$
• E. $\frac{25!}{(5!)^3}$

Hint:

Always divide by $n!$ when the groups are identical (unnamed) to avoid overcounting permutations of the groups.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This is a problem of partitioning a set of distinct objects into groups of equal size. There are two main steps: selecting the objects for each group, and then accounting for the fact that the groups themselves are not distinct (since they are of equal size).
Step 2: Key Formula or Approach:
The number of ways to divide \( (m \times n) \) distinct objects into \( n \) groups of size \( m \) each is given by: \[ \frac{(mn)!}{(m!)^n \cdot n!} \] The \( (m!)^n \) in the denominator is for the arrangements within each group (since we are forming groups, not ordered lists), and the \( n! \) is because the \( n \) groups of equal size are indistinguishable.
Step 3: Detailed Explanation:
Step 3a: Sequential Selection
First, let's select the objects for each group sequentially:
- Number of ways to choose 5 objects for the 1st group from 25: \( \binom{25}{5} \)
- Number of ways to choose 5 objects for the 2nd group from the remaining 20: \( \binom{20}{5} \)
- Number of ways to choose 5 objects for the 3rd group from the remaining 15: \( \binom{15}{5} \)
- Number of ways to choose 5 objects for the 4th group from the remaining 10: \( \binom{10}{5} \)
- Number of ways to choose 5 objects for the 5th group from the remaining 5: \( \binom{5}{5} \)
The total number of ways to form these ordered groups is the product: \[ \binom{25}{5} \binom{20}{5} \binom{15}{5} \binom{10}{5} \binom{5}{5} = \frac{25!}{20!5!} \cdot \frac{20!}{15!5!} \cdot \frac{15!}{10!5!} \cdot \frac{10!}{5!5!} \cdot \frac{5!}{0!5!} \] Canceling out the terms, this simplifies to: \[ \frac{25!}{5! \cdot 5! \cdot 5! \cdot 5! \cdot 5!} = \frac{25!}{(5!)^5} \] Step 3b: Accounting for Indistinguishable Groups
The calculation above assumes the groups are distinct (e.g., Group 1, Group 2, etc.). However, the problem states we are just "forming such groups," implying the groups are indistinguishable. Since there are 5 groups of the same size, we have overcounted by a factor of \( 5! \) (the number of ways to permute these 5 groups).
Therefore, we must divide our result by \( 5! \): \[ \text{Number of ways} = \frac{25!}{(5!)^5 \cdot 5!} \] Step 3c: Final Calculation
Combining the terms in the denominator: \[ \frac{25!}{(5!)^5 \cdot (5!)^1} = \frac{25!}{(5!)^6} \] Step 4: Final Answer:
The total number of ways of forming the groups is \( \frac{25!}{(5!)^6} \).