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Question Number 136448 by adhigenz last updated on 22/Mar/21

Mr.A wants to deliver 7 letters to his 7 friends so that each gets 1 letter.  All of the letters are written of the addresses of his 7 friends. Find the probbility that,  3 of his friends receive the correct letters and the remaining 4 receive the wrong ones.

$$\mathrm{Mr}.\mathrm{A}\:\mathrm{wants}\:\mathrm{to}\:\mathrm{deliver}\:\mathrm{7}\:\mathrm{letters}\:\mathrm{to}\:\mathrm{his}\:\mathrm{7}\:\mathrm{friends}\:\mathrm{so}\:\mathrm{that}\:\mathrm{each}\:\mathrm{gets}\:\mathrm{1}\:\mathrm{letter}. \\ $$$$\mathrm{All}\:\mathrm{of}\:\mathrm{the}\:\mathrm{letters}\:\mathrm{are}\:\mathrm{written}\:\mathrm{of}\:\mathrm{the}\:\mathrm{addresses}\:\mathrm{of}\:\mathrm{his}\:\mathrm{7}\:\mathrm{friends}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{probbility}\:\mathrm{that}, \\ $$$$\mathrm{3}\:\mathrm{of}\:\mathrm{his}\:\mathrm{friends}\:\mathrm{receive}\:\mathrm{the}\:\mathrm{correct}\:\mathrm{letters}\:\mathrm{and}\:\mathrm{the}\:\mathrm{remaining}\:\mathrm{4}\:\mathrm{receive}\:\mathrm{the}\:\mathrm{wrong}\:\mathrm{ones}. \\ $$

Answered by mr W last updated on 22/Mar/21

p=((P_3 ^7 ×!4)/(7!))=((!4)/(4!))=(9/(24))=(3/8)=37.5%

$${p}=\frac{{P}_{\mathrm{3}} ^{\mathrm{7}} ×!\mathrm{4}}{\mathrm{7}!}=\frac{!\mathrm{4}}{\mathrm{4}!}=\frac{\mathrm{9}}{\mathrm{24}}=\frac{\mathrm{3}}{\mathrm{8}}=\mathrm{37}.\mathrm{5\%} \\ $$

Commented by john_santu last updated on 23/Mar/21

i think it (1/(16)) sir

$${i}\:{think}\:{it}\:\frac{\mathrm{1}}{\mathrm{16}}\:{sir} \\ $$

Commented by john_santu last updated on 23/Mar/21

p= ((C_3 ^( 7)  ×!4)/(7!)) = ((7!.(!4))/(4!.3!.7!)) = (9/(24×6)) =(1/(16))

$${p}=\:\frac{{C}_{\mathrm{3}} ^{\:\mathrm{7}} \:×!\mathrm{4}}{\mathrm{7}!}\:=\:\frac{\mathrm{7}!.\left(!\mathrm{4}\right)}{\mathrm{4}!.\mathrm{3}!.\mathrm{7}!}\:=\:\frac{\mathrm{9}}{\mathrm{24}×\mathrm{6}}\:=\frac{\mathrm{1}}{\mathrm{16}} \\ $$

Commented by mr W last updated on 23/Mar/21

yes!

$${yes}! \\ $$

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