Question

There are

n

letters and

n

directed envelopes. The number of ways in which all the letters be put into wrong envelops?

Accepted Answer

Given, there are $n$ letters and $n$ directed envelopes. We need to find the no. of ways in which all the letters be put into wrong envelopes.

Now, total no. of ways of $n$ letters going to ang of the $n$ envelopes is $n!$

But, We need to subtract all arrangements where each letter goes to its directed envelope, which is equal to ${}^{n}c_{1} (n - 1)!$

So, no. of ways becomes $n! - {}^{n}c_{1} (n - 1)!$

But, again, we must add those arrangements where each set of two letters goes to their directed envelopes, as we subtracted then twice and no. of such arrangements are ${}^{n}c_{2} (n - 2)!$

So, no. of arrangements become $n! - {}^{n}c_{1} (n - 1)! + {}^{n}c_{2} (n - 2)!$

Again we must subtract those arrangements where each set of $3$ letters goes to their directed envelopes, because firstly we subtracted them and then added. Which are ${}^{n}c_{3} (n - 3)!$ .

So, this continues to given as a pattern, as $n! - {}^{n}c_{1} (n - 1)! + {}^{n}c_{2} (n - 2)! - {}^{n}c_{3} (n - 3)! + . . . . . . . . + {(- 1)}^{n} {}^{n}c_{n} (n - n)!$

$∵ {}^{n}c_{r} (n - r)! = \frac{n!}{r! (n - r)!} \cdot (n - r)! = \frac{n!}{r!}$

We can say, the required no. of ways is, $\{n! - \frac{n!}{1!} + \frac{n!}{2!} - \frac{n!}{3!} + . . . . . . . . + {(- 1)}^{n} \cdot \frac{n!}{n!}\}$

$= n! \{1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} . . . . . . . . + {(- 1)}^{n} \cdot \frac{1}{n!}\}$

or $= n! \{\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} . . . . . . . . + {(- 1)}^{n} \frac{1}{n!}\}$

This is also known as arrangement theorem.

There are $n$ letters and $n$ directed envelopes. The number of ways in which all the letters be put into wrong envelops?

Important Questions on Permutation and Combination

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