18 mice were placed in two experimental groups and one control group, with all groups equally large. In how ma

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Concept understanding — Permutations Without Repetition

Permutations Without Repetition – The Idea of Arranging Things

Imagine you have three different books on a shelf: a Physics book, a Chemistry book, and a Maths book. How many different ways can you arrange them in a row?

You could try listing them out:

Physics, Chemistry, Maths
Physics, Maths, Chemistry
Chemistry, Physics, Maths
Chemistry, Maths, Physics
Maths, Physics, Chemistry
Maths, Chemistry, Physics

That's 6 arrangements. Notice that each arrangement uses all three books exactly once — no book is repeated, and no book is left out. This is the core idea: permutations without repetition count the number of ways to arrange a set of distinct objects in order, using each object exactly once.

Why "Without Repetition"?

The phrase "without repetition" means that once you place an object in a position, you cannot use it again. In our book example, once you put the Physics book in the first slot, you cannot put it in the second or third slot. Each object appears exactly once in the arrangement.

This is different from "permutations with repetition" (like creating 3-letter codes from the letters A, B, C where you can reuse letters — e.g., AAA, AAB, etc.). Here, no repeats allowed.

The Counting Logic – Why Multiply?

Let's build the arrangement step by step for 3 distinct books:

First position: You have 3 choices (any of the 3 books).
Second position: After placing the first book, only 2 books remain — so 2 choices.
Third position: Only 1 book is left — so 1 choice.

Total arrangements = $3 \times 2 \times 1 = 6$ .

This product $3 \times 2 \times 1$ is called 3 factorial, written as $3!$ .

P(n) = n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1

For $n$ distinct objects, the number of permutations (arrangements in order) is $n!$ .

What If You Only Arrange Some of Them?

Suppose you have 5 different books, but you only want to arrange 3 of them on a shelf. How many ways?

First position: 5 choices
Second position: 4 choices
Third position: 3 choices

Total = $5 \times 4 \times 3 = 60$ .

This is a permutation of 5 objects taken 3 at a time, written as $P(5,3)$ or ${}^5P_3$ .

P(n, r) = \frac{n!}{(n-r)!} = n \times (n-1) \times \cdots \times (n-r+1)

Here $n$ is the total number of distinct objects, and $r$ is how many you are arranging. The formula works because:

Numerator $n!$ counts all arrangements of all $n$ objects.
Denominator $(n-r)!$ removes the arrangements of the $n-r$ objects you are not using.

Key Points to Remember

Important

Order matters — swapping two objects gives a different permutation.
No repetition — each object is used at most once.
For arranging all $n$ objects: $n!$
For arranging r out of $n$ objects: $\frac{n!}{(n-r)!}$

Common Mistake to Avoid …

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We're dividing 18 identical-role mice into three equal groups of 6; this is a partition problem where order among groups doesn't matter. The answer is $\frac{18!}{6! \cdot 6! \cdot 6! \cdot 3!}$ .

Why this isn't just "choose 6, then choose 6, then choose 6"

When you divide objects into groups where the groups themselves have no labels or distinguishing features, you're counting partitions, not arrangements. The mice are distinguishable (mouse 1, mouse 2, …, mouse 18), but the three groups—two experimental and one control—are described only by their sizes, not by which is "first" or "second."

The trap here is to compute $\binom{18}{6} \cdot \binom{12}{6} \cdot \binom{6}{6}$ and stop. That counts the number of ways to form three groups if the groups were labeled (say, Group A, Group B, Group C). But the problem says "two experimental groups and one control group"—the two experimental groups are interchangeable. We've overcounted by the number of ways to permute identical groups.

Step-by-step reasoning

If the groups were distinguishable, we'd pick 6 mice for the first group, then 6 from the remaining 12 for the second, then the last 6 automatically go to the third:

\binom{18}{6} \cdot \binom{12}{6} \cdot \binom{6}{6} = \frac{18!}{6! \cdot 12!} \cdot \frac{12!}{6! \cdot 6!} \cdot 1 = \frac{18!}{6! \cdot 6! \cdot 6!}.

Account for indistinguishable groups. The two experimental groups are identical in role; any partition that swaps them is the same division. We've counted each true partition $2! = 2$ times (once for each assignment of the two experimental labels). The control group is unique, so no further symmetry there.
Divide out the overcounting:

$\frac{18!}{6! \cdot 6! \cdot 6! \cdot 2!}.$ …

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Q.181818 mice were placed in two experimental groups and one control group, with all groups equally large. In how many ways can the mice be placed into three groups?

Why this isn't just "choose 6, then choose 6, then choose 6"

Step-by-step reasoning

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Q. $18$ mice were placed in two experimental groups and one control group, with all groups equally large. In how many ways can the mice be placed into three groups?

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