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July 1st, 2015, 09:42 AM   #1
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Question simple permutation and combination.

Let n be the number of ways in which 5 boys and 5 girls could stand such that all the girls stand consecutively. Let m be the number of ways in which 5 boys and 5 girls could stand such that exactly 4 girls stands consecutively. Then the ratio n/m will be ?

Please solve it, describing all the steps.

Last edited by skipjack; July 1st, 2015 at 03:59 PM.
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July 1st, 2015, 09:51 AM   #2
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If you have visited here but unable to solve it, then please type "no". It's so much annoying to have an unanswered question for hours.

Last edited by skipjack; July 1st, 2015 at 04:00 PM.
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July 1st, 2015, 12:16 PM   #3
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So far you have posted this same problem twice. The first time you called it "advanced". The second time you called it "simple".

Jonah said it right. Post what you have done. There are many people on this site who can help (not do it for you - help).

Both times you posted this question you used the words "permutation" and "combination". Use them.
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July 1st, 2015, 10:00 PM   #4
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Dear mrtwhs

every post in this forum is a request in itself. I am not ordering others.
helps are help.
And why I asked peoples to type "no" is because there were no activity for long and I wondered whether peoples are able to access my question or not.

And for the question. I have done
m= 5! x 6! x 5! [five girls can arrange in 5! ways then the group taken as one could arrange in between boys in 6! ways and those boys arrange themselves in 5! ways.]
n= 5C4 x 4! x 5! x 6! . [Four girls out of five are taken. They arrange in 4! ways. The remaining six persons can arrange in 6! ways. But the group of girls taken one can get 5 places, so they get there in 5! ways.]

I know it's incorrect, so I request for help.

Last edited by skipjack; July 1st, 2015 at 10:34 PM.
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July 2nd, 2015, 06:04 AM   #5
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Quote:
Originally Posted by rexdex View Post
Original post:
Let n be the number of ways in which 5 boys and 5 girls could stand such that all the girls stand consecutively. Let m be the number of ways in which 5 boys and 5 girls could stand such that exactly 4 girls stands consecutively. Then the ratio n/m will be ?

Your attempted solution:
m= 5! x 6! x 5! [five girls can arrange in 5! ways then the group taken as one could arrange in between boys in 6! ways and those boys arrange themselves in 5! ways.]
n= 5C4 x 4! x 5! x 6! . [Four girls out of five are taken. They arrange in 4! ways. The remaining six persons can arrange in 6! ways. But the group of girls taken one can get 5 places, so they get there in 5! ways.]
Well the first thing to notice is that the m and n in your original post are reversed from your attempted solution. So I'm going to choose one way and provide a hint:

m = the number of ways that 5 boys and 5 girls can line up so that the girls are together (the boys may be separated).

Your answer for this (5! x 6! x 5!) is too big by a factor of 5!. First, line up the 5 girls (5!) and then line up the six objects boy1, boy2, boy3, boy4, boy5, and the girls (6!). So the answer for m is m = 5! x 6!

See if you can now get n. Your final answer should be m/n = 1/5.
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July 2nd, 2015, 06:52 AM   #6
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we could chose 4 girls out of 5 in 5c4 ways which arrange in 4! ways and then we arrange the six objects in 7! ways subtracting the possibility of all the girls together.

so n would be. 5c4*4!*7! - ( situation of all girls to be together )

i have posted this question cause iam not able to solve. and not for fun or to test others. as mr mrt thinks.

however i thank him for giving hint for how to find m of my question.

iam not good in this part of algebra.

so i request help. for now finding n.
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July 2nd, 2015, 08:00 AM   #7
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Quote:
Originally Posted by rexdex View Post
we could chose 4 girls out of 5 in 5c4 ways which arrange in 4! ways and then we arrange the six objects in 7! ways subtracting the possibility of all the girls together.

so n would be. 5c4*4!*7! - ( situation of all girls to be together )
No need for subtraction. Line up the boys (5!) and then space them out like this:

_b1_b2_b3_b4_b5_

There are six blank spots. Pick one of them to put the group of 4 girls then pick a different one to put the solo girl.
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July 2nd, 2015, 08:25 AM   #8
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so. five boys are arranged in 5! ways and four girls out of five 5c4 ways. then the group can be placed in six places in 6! factorial ways . and the last girl in remaining 5 places in 5! ways
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July 2nd, 2015, 09:29 AM   #9
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Quote:
Originally Posted by rexdex View Post
so. five boys are arranged in 5! ways and four girls out of five 5c4 ways. then the group can be placed in six places in 6! factorial ways . and the last girl in remaining 5 places in 5! ways
Close but not quite. 5c4 just chooses the girls. You then need 4! to line them up. Once lined up, you can put them into any one of the 6 blank spots in 6 different ways (not 6!). Then put the last girl in a different spot.
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July 2nd, 2015, 09:51 AM   #10
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does the last girl goes in remaining 5 places in 5 ways?
will i have to multiply everything including 5 for the last girl?

thankyou for this much help. iwas stuck invthis problem from yesterday.
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