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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)

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

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

$$\mathrm{diarrangement}\:\mathrm{of}\:\mathrm{4}\:\mathrm{letters}\:=\:!\mathrm{4} \\ $$$$\mathrm{3}\:\mathrm{letters}\:\mathrm{are}\:\mathrm{in}\:\mathrm{correct}\:\mathrm{address} \\ $$$$\mathrm{this}\:\mathrm{3}\:\mathrm{letters}\:\mathrm{can}\:\mathrm{be}\:\mathrm{choose}\:\mathrm{in}\:\mathrm{C}_{\mathrm{3}} ^{\mathrm{7}} \:\mathrm{ways} \\ $$$$\therefore\:\mathrm{The}\:\mathrm{required}\:\mathrm{probability} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{!\mathrm{4}×\mathrm{C}_{\mathrm{3}} ^{\mathrm{7}} }{\mathrm{7}!}\:=\:\frac{\mathrm{1}}{\mathrm{16}} \\ $$$$\:\:\:\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{check}} \\ $$

Commented by mr W last updated on 21/Sep/20

correct!

$${correct}! \\ $$

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

thank you sir.

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by mr W last updated on 21/Sep/20

please try a similar question: Q111906

$${please}\:{try}\:{a}\:{similar}\:{question}: \\ $$$${Q}\mathrm{111906} \\ $$

Commented by john santu last updated on 21/Sep/20

what definition of !4 ? and what formula !n

$${what}\:{definition}\:{of}\:!\mathrm{4}\:? \\ $$$${and}\:{what}\:{formula}\:!{n}\: \\ $$

Commented by PRITHWISH SEN 2 last updated on 22/Sep/20

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

$$!\boldsymbol{\mathrm{n}}\:=\:\boldsymbol{\mathrm{n}}!\left[\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}!}+\frac{\mathrm{1}}{\mathrm{2}!}−\frac{\mathrm{1}}{\mathrm{3}!}+.......+\left(−\mathrm{1}\right)^{\mathrm{n}} \frac{\mathrm{1}}{\mathrm{n}!}\right] \\ $$

Commented by mr W last updated on 27/Sep/20

!n is the number of derangements from n elements.

$$!{n}\:{is}\:{the}\:{number}\:{of}\:{derangements} \\ $$$${from}\:{n}\:{elements}. \\ $$

Commented by PRITHWISH SEN 2 last updated on 22/Sep/20

sir please check Q 111906

$$\mathrm{sir}\:\mathrm{please}\:\mathrm{check}\:\mathrm{Q}\:\mathrm{111906} \\ $$

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)) =

$$ \\ $$$$\:\:^{\mathrm{7}} \mathrm{C}_{\mathrm{3}} ×\mathrm{4}!\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}!}+\frac{\mathrm{1}}{\mathrm{2}!}−\frac{\mathrm{1}}{\mathrm{3}!}+\frac{\mathrm{1}}{\mathrm{4}!}\right)\:\boldsymbol{\div}\:\mathrm{7}! \\ $$$$\:=\:\mathrm{35}×\mathrm{9}\boldsymbol{\div}\mathrm{5040}\:=\:\frac{\mathrm{1}}{\mathrm{16}} \\ $$$$\:=\: \\ $$

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