Letter arrangements: understanding probability and combinats

Counting 1 correct: 4C1=4, choosing which letter will be put in correct envelope. (Let's assume A letter was put in correct envelope).
Then 3 letters (BCD) and 3 envelopes (BCD) are left. # of ways to put them incorrectly is 2:
Envelopes: B-C-D
Letters: C-D-B
OR: D-B-C
So total # of ways one letter will be put into the envelope with its correct address is: 4C1*2=8. As total # of ways to distribute 4 letter in 4 envelopes is 4!=24, probability P(C=1)=8/24.

Counting 2 correct: 4C2=6, choosing which 2 letter will be put in correct envelope. (Let's assume A and B letter were put in correct envelopes).
Then 2 letters (CD) and 2 envelopes (CD) are left. # of ways to put them incorrectly is 1:
Envelopes: C-D
Letters: D-C
So total # of ways two letters will be put into the envelope with its correct address is: 4C2*1=6. As total # of ways to distribute 4 letter in 4 envelopes is 4!=24, probability P(C=2)=6/24.

Counting 3 correct: if three letters will be put in correct envelopes, then the fourth one also gets its correct envelope. So there is no way we can put exactly 3 letters in correct envelopes --> P(C=3)=0

Counting 4 correct: there are 4!=24 ways to distribute 4 letters in 4 envelopes and obviously only one is when all letters get their correct envelopes. P(C=4)=1/24.

Counting all incorrect, or 0 correct:
P(all incorrect)=1-(1 correct)-(2 correct)-(3 correct)-(4 correct)=1-8/24-6/24-0-1/24=9/24.




Opposite of "all letters in correct envelopes" (= 1-1/24 = 23/24), is: "at least one in wrong envelope" = "3 correct" (=0) + "2 correct" (=6/24) + "1 correct" (=8/24) + "0 correct, option A" (=9/24) --> \(1-\frac{1}{24}=\frac{23}{24}=0+\frac{6}{24}+\frac{8}{24}+\frac{9}{24}=\frac{23}{24}\).

Hope it's clear.