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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 (JEE Adv. 2014)
- a)264
- b)265
- c)53
- d)67
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Six cards and six envelopes are numbered 1, 2, 3, 4, 5, 6 and cards ar...
∵ Car d 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
Case I: Card numbered 2 is placed in envelope numbered 1.
Then it is dearrangement of 4 objects, which can be done in
= 9 waysCase II: Card numbered 2 is not placed in envelope numbered 1.
Then it is dearrangement of 5 objects, which can be done in
Then it is dearrangement of 5 objects, which can be done in
= 44 ways∴ Total ways = 44 + 9 = 53
Most Upvoted Answer
Six cards and six envelopes are numbered 1, 2, 3, 4, 5, 6 and cards ar...
To solve this problem, we can use the concept of derangements. A derangement is a permutation of the elements of a set, such that no element appears in its original position.
Let's denote the number of ways to arrange the remaining 5 cards (excluding card 1) in the envelopes as D(5). We can calculate D(5) using the following recursive formula:
D(n) = (n-1) * (D(n-1) + D(n-2))
where D(n) represents the number of derangements of n elements.
Now, let's calculate D(5):
D(5) = (5-1) * (D(4) + D(3))
= 4 * (D(4) + D(3))
= 4 * ((4-1) * (D(3) + D(2)) + (3-1) * (D(2) + D(1)))
= 4 * (3 * (D(3) + D(2)) + 2 * (D(2) + D(1)))
= 4 * (3 * ((3-1) * (D(2) + D(1)) + (2-1) * (D(1) + D(0))) + 2 * (D(2) + D(1)))
= 4 * (3 * (2 * (D(2) + D(1)) + 1 * (D(1) + D(0))) + 2 * (D(2) + D(1)))
= 4 * (3 * (2 * (1 * (D(1) + D(0))) + 1 * (D(1) + D(0))) + 2 * (D(2) + D(1)))
= 4 * (3 * (2 * (1 * (1 + 0)) + 1 * (1 + 0)) + 2 * (D(2) + D(1)))
= 4 * (3 * (2 * (1 * 1) + 1 * 1) + 2 * (D(2) + D(1)))
= 4 * (3 * (2 * 1 + 1 * 1) + 2 * (D(2) + D(1)))
= 4 * (3 * (2 + 1) + 2 * (D(2) + D(1)))
= 4 * (3 * 3 + 2 * (D(2) + D(1)))
= 4 * (9 + 2 * (D(2) + D(1)))
= 4 * (9 + 2 * (1 + D(0)))
= 4 * (9 + 2 * (1 + 1))
= 4 * (9 + 2 * 2)
= 4 * (9 + 4)
= 4 * 13
= 52
Therefore, there are 52 ways to arrange the remaining 5 cards in the envelopes. Since card 1 is always placed in envelope 2, there are 52 possible arrangements for the remaining 5 cards. Finally, we can place card 1
Let's denote the number of ways to arrange the remaining 5 cards (excluding card 1) in the envelopes as D(5). We can calculate D(5) using the following recursive formula:
D(n) = (n-1) * (D(n-1) + D(n-2))
where D(n) represents the number of derangements of n elements.
Now, let's calculate D(5):
D(5) = (5-1) * (D(4) + D(3))
= 4 * (D(4) + D(3))
= 4 * ((4-1) * (D(3) + D(2)) + (3-1) * (D(2) + D(1)))
= 4 * (3 * (D(3) + D(2)) + 2 * (D(2) + D(1)))
= 4 * (3 * ((3-1) * (D(2) + D(1)) + (2-1) * (D(1) + D(0))) + 2 * (D(2) + D(1)))
= 4 * (3 * (2 * (D(2) + D(1)) + 1 * (D(1) + D(0))) + 2 * (D(2) + D(1)))
= 4 * (3 * (2 * (1 * (D(1) + D(0))) + 1 * (D(1) + D(0))) + 2 * (D(2) + D(1)))
= 4 * (3 * (2 * (1 * (1 + 0)) + 1 * (1 + 0)) + 2 * (D(2) + D(1)))
= 4 * (3 * (2 * (1 * 1) + 1 * 1) + 2 * (D(2) + D(1)))
= 4 * (3 * (2 * 1 + 1 * 1) + 2 * (D(2) + D(1)))
= 4 * (3 * (2 + 1) + 2 * (D(2) + D(1)))
= 4 * (3 * 3 + 2 * (D(2) + D(1)))
= 4 * (9 + 2 * (D(2) + D(1)))
= 4 * (9 + 2 * (1 + D(0)))
= 4 * (9 + 2 * (1 + 1))
= 4 * (9 + 2 * 2)
= 4 * (9 + 4)
= 4 * 13
= 52
Therefore, there are 52 ways to arrange the remaining 5 cards in the envelopes. Since card 1 is always placed in envelope 2, there are 52 possible arrangements for the remaining 5 cards. Finally, we can place card 1
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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 (JEE Adv. 2014)a)264b)265c)53d)67Correct answer is option 'C'. Can you explain this answer?
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