Number of methods to post 4 letters in 4 envelopes such that only 1 letter goes to the correct envel

ABCD(envelopes)
ACDB(letters)
ADBC(letters)
(When A is in the right envelope other three have only 2 possible incorrect arrangements)
As we have 4 letters, total # of ways 4*2=8

Choosing right letter-right envelope can be done in 4C1 ways.
In each of the ways, for the remaining 3 letters and 3 envelopes, there are only 2 ways of putting all three in wrong envelopes as follows:
First one can be put in 2 wrong envelopes. Second and third can only be put in 1 remaining wrong envelope. So 2 ways.
Total 4C1*2 ways.
or 8 ways thus option 6 is correct

Total # of ways of choosing envelopes=4!=24.
Only one letter in the right envelope: 4(# of envelopes)*2(# of ways possible to arrange 3 letters incorrectly in the envelopes, when one is correct)
ABCD(envelopes)
ACDB(letters)
ADBC(letters)
(When A is in the right envelope other three have only 2 possible incorrect arrangements)
As we have 4 letters, total # of ways 4*2=8

hence the ans is 8...

one letter out of 4 that must be put in its correct envelope. We can do this in 4 ways. In how many ways can we put the rest of the 3 letters into 3 envelopes such that all 3 letters go into incorrect envelopes? We found that 3 letters can be put into 3 envelopes such that each letter is put in incorrectly in 2 ways. So total number of ways such that only one letters go into the envelopes with their correct addresses = 4*2 = 8 ways

one letter out of 4 that must be put in its correct envelope. We can do this in 4 ways.
rest of the 3 letters into 3 envelopes such that all 3 letters go into incorrect envelopes
so all ways is 4*2=8

1,2,3,4 and we have to set only 1 letter to correct envelope...so...
suppose 1 is in correct envelope...so remains are different.... brackets are in correct envelope
(1),(2),(3),(4) -> wrong
(1),(2), 4 , 3 -> wrong
(1), 3 , 2 ,(4) -> wrong
(1), 3 , 4 , 2 -> right .......
(1), 4 ,(3), 2 -> wrong
(1), 4 , 2 , 3 -> right .......
...
For 1 set there are 2 possibility are right...but we have 4 set ...
2*4 = 8
so, ans is 6)8

A B C D submit in a four 'a b c d' envelopes
if you fix A-----a
the other possibility in another envelopes is
b c d
----- ---- ----
C D B
D B C
so there is two possiblities if you fix the A
if you B,C,D totally 2*4==8.

Ans is 8

In 8 ways we can post letters to envolop such that only 1 letter goes to the correct envelope.

Consider => L1-e1 is right => then L2 has 2 possible combinations such as e3 or e4 (we consider L2-e3); it cannot take e2 bcoz then there will be a combination with 2 letters in right envelop. Now the obvious choice for L3 to take e4 and not e3 (same logic for L2) and thus L4 is paired bydefault with e2.....Now plz refer below sequence with 2 possibilities

Hence => [L1-e1 | L2-e3 | L3-e4 | L4-e2] OR [L1-e1 | L2-e4 | L3-e2 | L4-e3]

Now this sequence is only for L1-e1 to be the only right and all others are wrong; similarly we have 4 pairs such as (L1-e1 | L2-e2 | L3-e3 | L4-e4 ) with 2 possibilities for each

Hence Ans is 4 x 2 = 8 ways

Let’s first select the one letter out of 4 that must be put in its correct envelope. We can do this in 4 ways. In how many ways can we put the rest of the 3 letters into 3 envelopes such that all 3 letters go into incorrect envelopes? We found that 3 letters can be put into 3 envelopes such that each letter is put in incorrectly in 2 ways. So total number of ways such that only two letters go into the envelopes with their correct addresses = 4*2 = 8 ways

Correct letter can go to any one of the 4 envelopes.
Balance three letters can be placed in 2 ways such that none of these 3 letters go in correct envelope.
so total possible methods to post 4 letters in 4 envelopes such that only 1 letter goes to the correct envelope = 4*2=8

Those possibilites are...
ADBC
ACDB
CBDA
DBAC
BDCA
DACB
BCAD
CABD

So,there are 8 possibilities(Option 6).

let 1 letter and 1 envelope make together
then remaining arrangement has only 2 chances for each envelope so 2*2*2=8

l1 l2 l3. l4

l1 l3 l4 l2
l1 l4 l2 l3. 2 ways for l1
2 ways for l2 ,l3, l4 each
so 8