Question

There are six letters ${{L}_{1}},{{L}_{2}},{{L}_{3}},{{L}_{4}},{{L}_{5}},{{L}_{6}}$ and their corresponding six envelopes ${{E}_{1}},{{E}_{2}},{{E}_{3}},{{E}_{4}},{{E}_{5}},{{E}_{6}}$ . Letters having odd value can be put into odd value envelopes and even value letters can be put into even value envelopes, so that no letter goes into the right envelopes, then the number of arrangements equals.

Answer 1

Hint: The question is based on derangement. There are two sets of letters, odd numbered and even numbered. Now use the formula that the number of ways in which ‘n’ letters can be put in ‘n’ corresponding envelopes such that no letter goes to correct envelope is $n!\left( 1-\dfrac{1}{1!}+\dfrac{1}{2!}-\dfrac{1}{3!}.........+{{\left( -1 \right)}^{n}}\dfrac{1}{n!} \right)$ for both sets separately and multiply the results to get the answer.

Complete step by step solution:
The question is based on derangement. There are two sets of letters, odd numbered and even numbered. Let us first solve for the odd numbered letters.
So, there are three odd numbered letters and corresponding to each there are 3 odd numbered envelopes. We know that the number of ways in which ‘n’ letters can be put in ‘n’ corresponding envelopes such that no letter goes to correct envelope is $n!\left( 1-\dfrac{1}{1!}+\dfrac{1}{2!}-\dfrac{1}{3!}.........+{{\left( -1 \right)}^{n}}\dfrac{1}{n!} \right)$ . Therefore, the ways of putting odd number letters in odd numbered envelope such that no one is in the correct envelope is:
$3!\left( 1-\dfrac{1}{1!}+\dfrac{1}{2!}-\dfrac{1}{3!} \right)=6\left( 1-1+\dfrac{1}{2}-\dfrac{1}{6} \right)=6\times \dfrac{1}{3}=2$
Also, the number of derangement of 3 even numbered letter in even numbered envelopes is the same, i.e., $3!\left( 1-\dfrac{1}{1!}+\dfrac{1}{2!}-\dfrac{1}{3!} \right)=6\left( 1-1+\dfrac{1}{2}-\dfrac{1}{6} \right)=6\times \dfrac{1}{3}=2$ .
So, as both the above mentioned sets are to be put together simultaneously, such that letters having odd value can be put into odd value envelopes and even value letters can be put into even value envelopes, so that no letter go into the right envelopes, we take the product of the results of the two sets we solve above.
$2\times 2=4$
Therefore, the answer to the above question is 4.
Note: In questions related to derangement, there are only two possibilities to solve, either you count the cases manually else use the formula of derangement as we did in the above solution. Also, remember, whenever you apply the formula of derangement, the result of the formula is always a positive integer and can never be a fraction.