Q.One urn contains two black balls (labelled $B_1$ and $B_2$) and one white ball. A second urn contains one black ball and two white balls (labelled $W_1$ and $W_2$). Suppose the following experiment is performed. One of the two urns is chosen at random. Next a ball is randomly chosen from the urn. Then a second ball is chosen at random from the same urn without replacing the first ball.

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 $3!$.

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 ${}^5P_3$.

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

Common Mistake to Avoid …