AP EAMCET · Maths · Permutation Combination

A person writes letters to $6$ friends and addresses the corresponding envelopes. In how many ways can the letters be placed in the envelopes so that at least two of them are in the wrong envelopes?
Notation $: D_{n} = n! (\sum_{i = 0}^{n} \frac{{(- 1)}^{i}}{i!})$

A ${}^{6}C_{4} \cdot D_{2}$
B $\sum_{r = 3}^{6} {}^{6}C_{6 - r} \cdot D_{r}$
C $\sum_{r = 2}^{6} {}^{6}C_{6 - r} \cdot D_{r}$
D ${}^{6}C_{1} D_{5} + {}^{6}C_{0} . D_{6}$

Verified Solution

Answer & Solution

Correct Answer

(C) $\sum_{r = 2}^{6} {}^{6}C_{6 - r} \cdot D_{r}$

Step-by-step Solution

Detailed explanation

The number of ways in which at least two of the letters are in the wrong envelopes is given by ∑r=26C6-r6Dr=C46D2+C36D3+C26D4+C16D5+C06D6 where Dr=r!∑i=0r-1ii!

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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A person writes letters to 6 friends and addresses the corresponding envelopes. In how many ways can the letters be placed in the envelopes so that at least two of them are in the wrong envelopes?Notation :Dn=n!∑i=0n-1ii!

Step-by-step Solution

A person writes letters to $6$ friends and addresses the corresponding envelopes. In how many ways can the letters be placed in the envelopes so that at least two of them are in the wrong envelopes?
Notation $: D_{n} = n! (\sum_{i = 0}^{n} \frac{{(- 1)}^{i}}{i!})$