6 cards and 6 envelops 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
Answers
ANSWER:--------------
:D4+4∗(D3+3∗(D2+2∗(D1+1∗D0)))=9+4∗
(2+3∗(1+2∗(0+1)))=9+4∗(2+3∗(1+2))=9+4∗(2+9)=9+4∗11=53
More formally:
{we have n matching cards and enveloe numbers, and 1 isolated card and 1 isolated envelope.}
{ Let us say that Fn is the number of ways we can put the cards into the envelopes so that no numbers match. }
{Then we have in general:}
{F0=1(If you just have 1 isolated card and 1 envelope, put the card in the envelope)}
{Fn+1=Dn+1+(n+1)∗Fn(You can either put the isolated card into the isolated envelope, }
{giving a classic derangement problem for n+1, or put the isolated card into one of the other n+1 other envelopes, giving us Fnpossibilities for each)
hope it helps:--
T!—!ANKS!!!!
Answer:
Step-by-step explanation:
Total number of arrangements of 6 cards in 6 envelopes =6!
number of arrangements as card 1 in envelope 1 = 5!
number of arrangements as card 2 in envelope 2 = 5!
number of arrangements as card 3 in envelope 3 = 5! etc
number of arrangements as one card in one envelope = 6[5! ]
number of arrangements as 2 card in correct envelope = 6C2[4 ]!
] number of arrangements as 3 card in correct envelope = 6C3[3! ]
number of arrangements as 4 card in correct envelope = 6C4[2! ]
number of arrangements as 5 card in correct envelope = 6C5[1! ]
number of arrangements as 6 card in correct envelope = 6C6[0! ]
WE NOW CALCULATE
number of arrangements all card in incorrect envelope
6! -6[5!] +6C2[4! ] -6C3[3! ] +6C4[2!] -6C5[1! ] +6C6[0! ]
= 720 - 6[120] +15[24] -20[6] +15[2] -6[1] +1[1]
=720 -720 +360 -120 +30-6+1
=265
but we want to find number of arrangements all card in incorrect envelope and also card 1 in envelope 2
in this 265 arrangements card 1 may be in
envelope 2 or 3 or 4 or 5 or 6 [ total five chances ]
so number of such arrangements with card 1 in envelope 2
= 265/5 =53