5 girls and 3 boys are arranged randomly in a row

Question

5 girls and 3 boys are arranged randomly in a row. Find the probability that:

1) there is one boy on each end.
2) There is one girl on each end.

A. 3/28 & 5/14
B. 5/28 & 3/14
C. 5/14 & 3/28
D. 3/28 & 1/14
E. 1/28 & 5/14

Bunuel · Accepted Answer

Total # of arrangements is 8!;

1. The # of arrangements where there is one boy on each end is  C^2_3*2*6! , where  C^2_3  is # of ways to select which 2 boys will be on that positions, *2 # of way to arrange them on that position (A------B and B------A) and 6! is # of arrangements of 6 people left;

P=\frac{C^2_3*2*6!}{8!}=\frac{3}{28} .

2. The # of arrangements where there is one girl on each end is  C^2_5*2*7! , where  C^2_5  is # of ways to select which 2 girls will be on that positions, *2 # of way to arrange them on that position (A------B and B------A) and 6! is # of arrangements of 6 people left;

P=\frac{C^2_5*2*6!}{8!}=\frac{5}{14} .

Answer: A.

Bunuel · Answer

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swyw · Answer

Hi Bunel, firstly I would like to thank you for all the help you've given to the forum! You have been nothing but superb!

I have a question on this though, would it be possible also to solve the question as below?
I had solved it like this initially and I'm not sure now if I have some mistakes in my fundamental theory.
It would be great if you could help clarify.
Thank you!

1)
 P=(P^2_3*6!)/8!

2)
 P=(P^2_5*6!)/8!

Bunuel · Answer

Yes, your approach is fine: 3P2=3C2*2! and 5P2=5C2*2!.

Hope it's clear.

LalaB · Answer

one boy on each end = (3/8)*(2/7)=3/28
one girl on each end =(5/8)*(4/7)=5/14

skamal7 · Answer

bunnel,
 Any problem of similar type available in this forum??? if so can you please tag those ..thanks in advance

pharmademic · Answer

Can you please show the math for  C^2_3 ?

Bunuel · Answer

3C2 = \frac{3!}{2!*(3-2)!}

21. Combinatorics/Counting Methods 
     Theory   
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For more:
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Hope it helps.

nick1816 · Answer

We have restrictions only for the seats that are at the end. It doesn't matter who sits on the seats which are not at the end.

Number of ways students can sit on the end seats without any constraints= 8*7=56

Number of ways students can sit on the end seats, when there is one boy on each end= 3*2=6
Probability that there is one boy on each end= 6/56=3/28

Number of ways students can sit on the end seats, when there is one girl on each end = 5*4=20
Probability that there is one girl on each end= 20/56=5/14

sanjna2023 · Answer

Another approach using FCP - 
Total arrangements = 8!
Arranging 6 people in between 2 boys/ girls = 6!
1. B - 6! - B (Format)

3 * 2 * 6!/ 8! = 3/28

2. G - 6! - G
5*4*6! = 5/14

KeyurJoshi · Answer

Total possible cases with 5 girls and 3 boys = 8!

B _ _ _ _ _ _ B

So to select 2 boys for each end and these boys can be at either ends so = 3C2 * 2! = 6
AND possible combination for middle seats i.e for 1 boy and 5 girls = total 6!

Probability = Boys at either ends / Total possibilities = 6 * 6! / 8! = 3/28
 
Similarly

G _ _ _ _ _ _ G

So to select 2 girls for each end and these girls can be at either end so = 5C2 * 2! = 20
AND possible combination for middle seats i.e for 3 boys and 3 girls = total 6!

Probability = Girls at either ends / Total possibilities = 20 * 6! / 8! = 5/14

MathRevolution · Answer

Girls: 5 and Boys: 3

Total: 8 => Total number of ways to arrange 8 people is 8!

One boy at each end: Select 2 boys out of 3:  ^3{C_2} . Now, these two can be arranged in 2! ways. The remaining 5 girls and 1 boy can be arranged in 6! ways.

=>  ^3{C_2}  * 2! * 6!

=> Probability:  \frac{(^3{C_2} * 2! * 6!) }{ 8!}  =  \frac{3 }{28}

One girl at each end: Select 2 girls out of 5:  ^5{C_2} . Now, these two can be arranged in 2! ways. The remaining 3 girls and 3 boys can be arranged in 6! ways.

=>  ^5{C_2}  * 2! * 6!

=> Probability:  \frac{(^5{C_2} * 2! * 6!) }{ 8!}  =  \frac{5 }{14}

Answer A

LorenzoUboldi · Answer

My approach:

1) B------B --> (6!/5!) / (8!/3!5!) = 3/28

2) G------G --> (6!/3!3!) / (8!/5!3!) = 5/14

Answer A

Foreheadson · Answer

Alternatively, since there is not need for boys and girls to be unique:
if you put 2 boys at ends, 1 boy and 5 girls are left to distribute among 6 slots, 6C1=6. and 5 girls and 3 boys between 8 places is 8C3=56. => 6/56/=3/28
if you put 2 girls at ends, 3 boys and 3 girls are left to distribute among 6 slots, 6C3=20. and 5 girls and 3 boys between 8 places is 8C3=56. => 20/56/=5/14

Kinshook · Answer

Given: 5 girls and 3 boys are arranged randomly in a row. 
Asked: Find the probability that:

1) there is one boy on each end.
Number of ways to arrange 5 girls and 3 boys such that there is one boy on each end = 3C2*2*6!
Total number of ways to arrange 5 girls and 3 boys = 8!
The probability that there is one boy on each end = 3C2*2*6!/8! = 3/28

2) There is one girl on each end.
Number of ways to arrange 5 girls and 3 boys such that there is one girl on each end = 5C2*2*6!
Total number of ways to arrange 5 girls and 3 boys = 8!
The probability that there is one boy on each end = 5C2*2*6!/8! = 5/14

IMO A

Arjun123321 · Answer

Thanks for the above explanation. I have a quick query.
Shouldnt we divide the total # of arrangements (8!) by 5! and 3!, since boys and girls are the only 2 elements (and not 8 different elements) and thus to avoid repetation of arrangements?

Bunuel · Answer

We apply a factorial correction for identical items. However, when considering people, as each individual is unique, this correction is not necessary; every permutation generated by 8! will be distinct from another.

bumpbot · Answer

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5 girls and 3 boys are arranged randomly in a row

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