If the letters of the word ASSASSINATION are arranged at random, find the probability that (a) Four S's come c

NCERT Exemplar · Q14

Q. If the letters of the word ASSASSINATION are arranged at random, find the probability that

(a) Four S's come consecutively in the word.

(b) Two I's and two N's come together.

(d) No two A's are coming together.

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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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ASSASSINATION has 13 letters (A $\times$ 3, S $\times$ 4, I $\times$ 2, N $\times$ 2, T $\times$ 1, O $\times$ 1), giving a total of $N=\dfrac{13!}{3!\,4!\,2!\,2!}=10{,}810{,}800$ equally likely arrangements. The required probabilities are (a) $\dfrac{2}{143}$ ,

(b) $\dfrac{1}{39}$ ,

(d) $\dfrac{15}{26}$ .

Setting up the total number of arrangements

The word ASSASSINATION has 13 letters made up of:

A : 3 times
S : 4 times
I : 2 times
N : 2 times
T : 1 time
O : 1 time

Since some letters repeat, the number of distinct arrangements is given by the permutations-with-repetition formula:

N=\dfrac{13!}{3!\,4!\,2!\,2!\,1!\,1!}=\dfrac{13!}{3!\,4!\,2!\,2!}

The denominator is $3!\,4!\,2!\,2!=6\times 24\times 2\times 2=576$ , and $13!=6{,}227{,}020{,}800$ , so

N=\dfrac{6{,}227{,}020{,}800}{576}=10{,}810{,}800.

This total $N=10{,}810{,}800$ is the denominator for every part below.

(a) The four S's come consecutively

Tie the four S's into one block (SSSS). We now arrange this block together with A, A, A, I, I, N, N, T, O — that is $1+3+2+2+1+1=10$ units, with A repeated 3 times and I, N repeated twice each:

\text{favourable}=\dfrac{10!}{3!\,2!\,2!}=\dfrac{3{,}628{,}800}{24}=151{,}200.

P(a)=\dfrac{151{,}200}{10{,}810{,}800}=\dfrac{2}{143}.

(Symbolically $P(a)=\dfrac{10!\,4!}{13!}=\dfrac{24}{13\times 12\times 11}=\dfrac{24}{1716}=\dfrac{2}{143}$ .)

(b) The two I's and the two N's come together

Tie the two I's into a block (II) and the two N's into a block (NN). We arrange these two blocks together with A, A, A, S, S, S, S, T, O — that is $1+1+3+4+1+1=11$ units, with A repeated 3 times and S repeated 4 times:

\text{favourable}=\dfrac{11!}{3!\,4!}=\dfrac{39{,}916{,}800}{144}=277{,}200.

P(b)=\dfrac{277{,}200}{10{,}810{,}800}=\dfrac{1}{39}.

(c) All the A's do not come together

Use the complement. First count the arrangements in which all three A's do stay together by tying them into one block (AAA). This block plus S, S, S, S, I, I, N, N, T, O gives $1+4+2+2+1+1=11$ units: …

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Q.If the letters of the word ASSASSINATION are arranged at random, find the probability that

Setting up the total number of arrangements

(a) The four S's come consecutively

(b) The two I's and the two N's come together

(c) All the A's do not come together

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