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A permutation of $1,2, \dots, n$ is chosen at random. Then the probability that the numbers $1$ and $2$ appear as neighbour equals

1. $\frac{1}{n}$
2. $\frac{2}{n}$
3. $\frac{1}{n-1}$
4. $\frac{1}{n-2}$

retagged | 87 views

## 1 Answer

+3 votes

Answer: $\mathbf B$

Explanation:

The number of integers are $\mathbf n$

Total number of arrangements possible $= \mathbf n!$

Now, let's bundle  $1$ and $2$ together, then number of arrangements possible $\mathrm{=(n-1)!2!}$ ways.

$\because$ Total elements now are $\color {blue}{\mathrm n-1}$ and one bundle is having $2$ elements which can be arranged in $2!$ ways.

So, favorable cases $\mathrm{=(n-1)!2!}$

So, total cases possible, i.e., sample space $=\mathrm n!$

$\therefore$ Probability $\large{\mathrm{= \frac{(n-1)!2!}{n!} = \frac{(n-1)!2!}{(n-1)!n}=\color {green} {\frac{2}{n} }}}$

$\therefore \mathbf B$ is the correct option.

by Boss (19.4k points)
edited by
+1
Nice explanatory solution.
+1
Thanks.
0
Why it's n-1? And not n-2?
+1
Because when you make group of $2$ then you will get $\mathbf{n-1}$ elements.

$\mathbf{Eg}$:

Let's say given elements are: $\{1,2,3,4,5\}$

Now you want to group $\{1,2\}$

So, total elements would now be:

$\{\enclose{circle}{\underset{{\text{count as 1}}}{1,2}},3,4,5\}$

So, total elements are :$5-1 = 4 = \mathbf{n-1}$

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