Question Number 114875 by bobhans last updated on 21/Sep/20

A man sent 7 letters to his 7 friend .  the letters are kept in addressed envelopes  at random. the probability that 3 friends  receive correct letters and 4 letters go  to wrong destination is _    (old question unanswered)

Answered by PRITHWISH SEN 2 last updated on 21/Sep/20

diarrangement of 4 letters = !4  3 letters are in correct address  this 3 letters can be choose in C_3 ^7  ways  ∴ The required probability                                            = ((!4×C_3 ^7 )/(7!)) = (1/(16))     please check

Commented bymr W last updated on 21/Sep/20

correct!

Commented byPRITHWISH SEN 2 last updated on 21/Sep/20

thank you sir.

Commented bymr W last updated on 21/Sep/20

please try a similar question:  Q111906

Commented byjohn santu last updated on 21/Sep/20

what definition of !4 ?  and what formula !n

Commented byPRITHWISH SEN 2 last updated on 22/Sep/20

!n = n![1−(1/(1!))+(1/(2!))−(1/(3!))+.......+(−1)^n (1/(n!))]

Commented bymr W last updated on 27/Sep/20

!n is the number of derangements  from n elements.

Commented byPRITHWISH SEN 2 last updated on 22/Sep/20

sir please check Q 111906

Answered by soumyasaha last updated on 22/Sep/20

    ^7 C_3 ×4!(1−(1/(1!))+(1/(2!))−(1/(3!))+(1/(4!))) ÷ 7!   = 35×9÷5040 = (1/(16))   =