Question

In how many ways 6 letters can be placed in 6 envelopes such that:
No letter is placed in its corresponding envelope?.
(a) 265
(b) 275
(c) 255
(d) None of the above.

Answer 1

We solve this problem by using the derangements formula. We are asked to find the number of ways of placing 6 letters in 6 envelopes such that no letter is placed in its corresponding letter which means that we need to do the derangements which means that we should not place the object in its place. The formula for the derangements of $'n'$ objects to $'n'$ places is given as
$n!\left( 1-\dfrac{1}{1!}+\dfrac{1}{2!}-\dfrac{1}{3!}+.....+{{\left( -1 \right)}^{n}}\dfrac{1}{n!} \right)$
By using the above formula we find the required number of ways.

Complete step-by-step solution
We are given that there are 6 letters with 6 envelopes.
We are asked to find the number of ways of arranging the letters in the envelopes such that no letter is placed in its corresponding envelope.
We know that the number of ways of arranging the objects in the respected places such that no object is placed in its corresponding places is nothing but the number of derangements.
So, we need to find the number of derangements of 6 letters in 6 envelopes.
Let us assume that the number of derangements of 6 letters in 6 envelopes as $'W'$
We know that the formula for the derangements of $'n'$ objects to $'n'$ places is given as
$n!\left( 1-\dfrac{1}{1!}+\dfrac{1}{2!}-\dfrac{1}{3!}+.....+{{\left( -1 \right)}^{n}}\dfrac{1}{n!} \right)$
Now, by using the above formula we get the value of $'W'$ as
$\Rightarrow W=6!\left( 1-\dfrac{1}{1!}+\dfrac{1}{2!}-\dfrac{1}{3!}+\dfrac{1}{4!}-\dfrac{1}{5!}+\dfrac{1}{6!} \right)$
We know that the formula of factorial that is
$\Rightarrow n!=1\times 2\times 3\times ......\times n$
By using the factorial formula in above equation we get
$\Rightarrow W=720\left( 1-1+\dfrac{1}{2}-\dfrac{1}{6}+\dfrac{1}{24}-\dfrac{1}{120}+\dfrac{1}{720} \right)$
Now, by multiplying the terms with 720 which is outside the brackets we get

& \Rightarrow W=360-120+30-6+1 \\\ & \Rightarrow W=265 \\\ \end{aligned}$$ **Therefore the number of ways of arranging the 6 letters in 6 envelopes in the given condition is 265. So, option (a) is the correct answer.** **Note:** Students may make mistakes in finding the number of ways in the general method without using the formula. They may do the solution as follows. We know that the number of ways of arranging the 6 letters in 6 envelopes without any condition is given as $$6!=720$$ Now, we know that the number of ways of arranging the 6 letters in 6 envelopes correctly can be done only in 1 way So, the total number of ways of arranging the 6 letters in 6 envelopes such that no letter is placed in its corresponding envelope is given as $$720-1=719$$ This solution is wrong because we are given that no letter should be placed in its corresponding envelope. But, the answer 719 ways includes the number of ways that at most 6 letters can be placed in its corresponding envelopes which is not the required answer.