Six cards and six envelopes are numbered 1, 2, 3, 4, 5, 6 and cards are to be placed in envelopes so that each envelope contains exactly one card and no card is placed in the envelope bearing the same number and moreover the card numbered 1 is always placed in envelope numbered 2. Then the number of ways it can be done is
A.
264
B.
265
C.
53
D.
67
Answer :
53
Solution :
$$\because $$ Card numbered 1 is always placed in envelope numbered 2, we can consider two cases. Case I : Card numbered 2 is placed in envelope numbered 1.
Then it is dearrangement of 4 objects, which can be done in $$4!\left( {1 - \frac{1}{{1!}} + \frac{1}{{2!}} - \frac{1}{{3!}} + \frac{1}{{4!}}} \right) = 9\,\,{\text{ways}}$$ Case II : Card numbered 2 is not placed in envelope numbered 1.
Then it is dearrangement of 5 objects, which can be done in $$5!\left( {1 - \frac{1}{{1!}} + \frac{1}{{2!}} - \frac{1}{{3!}} + \frac{1}{{4!}} - \frac{1}{{5!}}} \right) = 44\,\,{\text{ways}}$$
∴ Total ways = 44 + 9 = 53
Releted MCQ Question on Algebra >> Permutation and Combination
Releted Question 1
$$^n{C_{r - 1}} = 36,{\,^n}{C_r} = 84$$ and $$^n{C_{r + 1}} = 126,$$ then $$r$$ is:
Ten different letters of an alphabet are given. Words with five letters are formed from these given letters. Then the number of words which have at least one letter repeated are
Eight chairs are numbered 1 to 8. Two women and three men wish to occupy one chair each. First the women choose the chairs from amongst the chairs marked 1 to 4 ; and then the men select the chairs from amongst the remaining. The number of possible arrangements is