The Student Room Group

A boy and A girl standing next to each-other.

Suppose that 7 boys and 13 girls line up in a row.
Let S be the number of places in the row where a boy and a
girl are standing next to each other. For example, for the row
GBBGGGBGBGGGBGBGGBGG we have S = 12. Find the
average value of S (if all possible orders of these 20 people are
considered).

For i=1,2,.....,19. Let NiN_i be the number of times boy-girl or girl-boy pair occupy position i and i+1. (We are taking boys and girls to be distinguishable while calculating Ns. How is the average value of S 119Ni20!\frac{\sum_{1}^{19} N_i}{20!} ?

Also why would you get the same answer if you consider boys and girls to be indistinguishable from each other?
Original post by Quantum Horizon
Suppose that 7 boys and 13 girls line up in a row.
Let S be the number of places in the row where a boy and a
girl are standing next to each other. For example, for the row
GBBGGGBGBGGGBGBGGBGG we have S = 12. Find the
average value of S (if all possible orders of these 20 people are
considered).

For i=1,2,.....,19. Let NiN_i be the number of times boy-girl or girl-boy pair occupy position i and i+1. (We are taking boys and girls to be distinguishable while calculating Ns. How is the average value of S 119Ni20!\frac{\sum_{1}^{19} N_i}{20!} ?
For each permutation P of the 20 boys and girls, define Ni(P)N_{i}(P) to equal 1 if a boy-girl or girl-boy pair occupy positions i, i+1, and to equal 0 otherwise.

Then for any permutation P, the number of places where a boy + girl are standing next to each other is 119Ni(P)\sum_1^{19} N_i(P).

Summing and averaging over all permutations gives the desired result.

Also why would you get the same answer if you consider boys and girls to be indistinguishable from each other?
This is unclear: what do you consider the question to be in this case. (If you don't think the question has changed, obviously the answer won't changed).
Thanks for the reply.
Original post by DFranklin

This is unclear: what do you consider the question to be in this case. (If you don't think the question has changed, obviously the answer won't changed).


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If you look at the solution provided, the author assumes boys and girls to be indistinguishable. This gives the same answer as above where each NiN_i is 2*7*13*18! (2 for girl-boy or boy-girl, 7 for boys and 13 for the girls) despite the fact that we are using permutation and he is using combinations. And we know that there are no coincidences in combinatorics. I was wondering why is this the case?
Original post by Quantum Horizon
If you look at the solution provided, the author assumes boys and girls to be indistinguishable. This gives the same answer as above where each NiN_i is 2*7*13*18! (2 for girl-boy or boy-girl, 7 for boys and 13 for the girls) despite the fact that we are using permutation and he is using combinations. And we know that there are no coincidences in combinatorics. I was wondering why is this the case?
Well, if we have a random arrangement (boys indistinguishable, girls indisinguishable), such as:

BGGBGGBGGBGGBGGBGGBG (not terribly random, TBH, but it doesn't matter!)

then there are precisely 7!13! ways of doing this where you do distinguish the boys and girls. (7! ways of arranging the boys, 13! ways of arranging the girls).

So if you count the number of scenarios where the i_th position is BG or GB (i.e. the N_i of the question), then you can go from indistinguishable to distinguishable by multiplying by 7!13!.

But the total number of arrangements increases from 20C7 = 20!/(7!13!) to 20! as you go from indistinguishable to distinguishable, and so the average is unchanged.

Basically, what you're averaging doesn't care if the Boys are distinguisable, so in the end either method of counting gives the same result.

I feel it would be remiss to ignore the "probability" method of solving this question:

Consider the i, i+1 positions, define I_i = 1 if we have BG or GB in these positions. 0 otherswise. Now p(BG) + p(GB) = 2 * 7 * 13 / (20 * 19). So E[I_i] = 13 * 14 / (20 * 19).

By linearity of expectation, E[119Ii]=119E[Ii]=19×13×1420×19=13×14/20=91/10E[\sum_1^{19} I_i] = \sum_1^{19} E[I_i] = \dfrac{19 \times 13 \times 14 }{20 \times 19} = 13 \times 14 / 20 = 91 / 10
Original post by Quantum Horizon
Suppose that 7 boys and 13 girls line up in a row.
Let S be the number of places in the row where a boy and a
girl are standing next to each other. For example, for the row
GBBGGGBGBGGGBGBGGBGG we have S = 12. Find the
average value of S (if all possible orders of these 20 people are
considered).

For i=1,2,.....,19. Let NiN_i be the number of times boy-girl or girl-boy pair occupy position i and i+1. (We are taking boys and girls to be distinguishable while calculating Ns. How is the average value of S 119Ni20!\frac{\sum_{1}^{19} N_i}{20!} ?

Also why would you get the same answer if you consider boys and girls to be indistinguishable from each other?


**** dis math!@ WTF is this dishtiu! ?! DISHWASHERS!
Original post by Luke5125
**** dis math!@ WTF is this dishtiu! ?! DISHWASHERS!


Thank for your enlightening reply. I wonder what I would've done without you. Very helpful???
Original post by DFranklin
Well, if we have a random arrangement (boys indistinguishable, girls indisinguishable), such as:

BGGBGGBGGBGGBGGBGGBG (not terribly random, TBH, but it doesn't matter!)

then there are precisely 7!13! ways of doing this where you do distinguish the boys and girls. (7! ways of arranging the boys, 13! ways of arranging the girls).

So if you count the number of scenarios where the i_th position is BG or GB (i.e. the N_i of the question), then you can go from indistinguishable to distinguishable by multiplying by 7!13!.

But the total number of arrangements increases from 20C7 = 20!/(7!13!) to 20! as you go from indistinguishable to distinguishable, and so the average is unchanged.

Basically, what you're averaging doesn't care if the Boys are distinguisable, so in the end either method of counting gives the same result.

I feel it would be remiss to ignore the "probability" method of solving this question:

Consider the i, i+1 positions, define I_i = 1 if we have BG or GB in these positions. 0 otherswise. Now p(BG) + p(GB) = 2 * 7 * 13 / (20 * 19). So E[I_i] = 13 * 14 / (20 * 19).

By linearity of expectation, E[119Ii]=119E[Ii]=19×13×1420×19=13×14/20=91/10E[\sum_1^{19} I_i] = \sum_1^{19} E[I_i] = \dfrac{19 \times 13 \times 14 }{20 \times 19} = 13 \times 14 / 20 = 91 / 10


PRSOM! and Thank You!!
Original post by Quantum Horizon
Thank for your enlightening reply. I wonder what I would've done without you. Very helpful???


Ge, Thanks DUDE!
Reply 8
May I ask what level of math this is/is equivalent to? (as in: A2 stats, AS further, A2 further, etc or other)
I honestly thought that this was a heckpost but it's a genuine maths question lol

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