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In how many ways can 5 boys and 4 girls be arranged in a line so that

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In how many ways can 5 boys and 4 girls be arranged in a line so that  [#permalink]

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New post 16 Apr 2019, 20:47
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A
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C
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E

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Re: In how many ways can 5 boys and 4 girls be arranged in a line so that  [#permalink]

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New post 16 Apr 2019, 21:13
5
B _ _ _ _ _ _ _ B

The two boys can be selected in \(^5C_2\) ways and can be arranged in 2! ways. The remaining 7 people can be selected in 1 way and can be arranged in 7! ways.

\(\frac{5!}{2!*3!}*2! * 7!\)

\(\frac{5!}{3!}*7!\)

OPTION: C
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Re: In how many ways can 5 boys and 4 girls be arranged in a line so that  [#permalink]

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New post 16 Apr 2019, 22:47
Bunuel wrote:
In how many ways can 5 boys and 4 girls be arranged in a line so that there will be a boy at the beginning and at the end?


(A) 3!/5! * 7!

(B) 5!/6! * 7!

(C) 5!/3! * 7!

(D) 3!/5! * 7!

(E) 5!/7! * 7!


total arragements of boys
5c2 *2 ways and total places left after 2 boys occupy first and last position 7
so 5c2*2*7!
IMO C 5!/3! * 7!
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Re: In how many ways can 5 boys and 4 girls be arranged in a line so that  [#permalink]

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New post 22 Apr 2019, 02:45
Bunuel wrote:
In how many ways can 5 boys and 4 girls be arranged in a line so that there will be a boy at the beginning and at the end?


(A) 3!/5! * 7!

(B) 5!/6! * 7!

(C) 5!/3! * 7!

(D) 3!/5! * 7!

(E) 5!/7! * 7!


The way the answers are given , thought that 7! is in the denominator.

B-------B
5*7*6*5*4*3*2*1*4
That's option C
\(\frac{5!}{3!}*7!\)
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Re: In how many ways can 5 boys and 4 girls be arranged in a line so that  [#permalink]

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New post 10 Aug 2019, 02:03
Slight confusion.

When we say 7!, aren't we considering all 7 elements to be distinct ? Where as there are 3 Boys and 4 Girls. Why are we considering each boy to be different and each girl to be different ?

Will appreciate a response.
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Re: In how many ways can 5 boys and 4 girls be arranged in a line so that  [#permalink]

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New post 10 Aug 2019, 02:11
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Re: In how many ways can 5 boys and 4 girls be arranged in a line so that  [#permalink]

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New post 10 Aug 2019, 04:10
Bunuel wrote:
altairahmad wrote:
Slight confusion.

When we say 7!, aren't we considering all 7 elements to be distinct ? Where as there are 3 Boys and 4 Girls. Why are we considering each boy to be different and each girl to be different ?

Will appreciate a response.

______________________
Aren't all people different?


How can I determine if B1B2b3B4 are to be taken as 4 distinct objects or same ? I am comparing this scenario to ABCCADA. The letters are repeating, much like the boys and girls in this question. Why didn't we divide the Boys and Girls 7! by 4! X 3!.

Thanks againm

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Re: In how many ways can 5 boys and 4 girls be arranged in a line so that  [#permalink]

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New post 10 Aug 2019, 04:19
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Bunuel wrote:
In how many ways can 5 boys and 4 girls be arranged in a line so that there will be a boy at the beginning and at the end?


(A) 3!/5! * 7!

(B) 5!/6! * 7!

(C) 5!/3! * 7!

(D) 3!/5! * 7!

(E) 5!/7! * 7!


Number of ways of arranging 2 boys at the ends = 5p2 = 5!/3!

The remaining 7 people can be arranged in 7! Ways

So, total = 5!/3!*7!

Option C

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Re: In how many ways can 5 boys and 4 girls be arranged in a line so that   [#permalink] 10 Aug 2019, 04:19

In how many ways can 5 boys and 4 girls be arranged in a line so that

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