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Mathematics 21 Online
OpenStudy (anonymous):

7 men and 8 woman are going to be assigned seats in a movie theather. if 15 tickets with numbered seats are given at ramdom, find the prob that exaclty one man is in one of the first 5 seats

OpenStudy (anonymous):

35*210? Here's my reasoning: there are five ways any man can sit in the first five seats: MWWWW WMWWW WWMWW WWWMW WWWWM And there are seven men that can be that man, so 5*7 = 35 Now , the remaining ten seats have 6 men left to seat, which has this many arrangements: (10*9*8*7*6*5)/(6!) = 210 35 * 210 = 7350

OpenStudy (anonymous):

wronggg

OpenStudy (anonymous):

its suppose to be a fraction

OpenStudy (anonymous):

Oooooh, ok

OpenStudy (anonymous):

then it is 7350 out of 259459200

OpenStudy (anonymous):

so 7350/259459200

OpenStudy (anonymous):

well, it reduces

OpenStudy (anonymous):

147/5189184, which seems ridiculous

OpenStudy (anonymous):

is it right

OpenStudy (anonymous):

hold on...

OpenStudy (anonymous):

The total number of choices for people to sit is...

OpenStudy (anonymous):

is

OpenStudy (anonymous):

I don;t know why I can't get the total. I'll brb

OpenStudy (anonymous):

kk

OpenStudy (anonymous):

okay, try this, it doesn't matter which man it is. that was my mistake, so there aren't 7350 ways they can sit with one man in the first 5 seats, there are only 10C6*5 = 1050. then there are 15C7 total ways, for 6435 total 1050/6435 = 70/429, less ridiculous.

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