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Twenty children, some boys, some girls, sit at a round table. Each child wears a blue or a red T-shirt. The right neighbor of each boy wears a blue T-shirt, and the left neighbor of each girl wears a red T-shirt. How many boys can there be?

work done:

I have showed that only 10 boys or 1 boy by showing the pattern of $$bbg$$ cannot exist as it would mean the boy in the middle would wear a blue and red shirt which can't be . I am trying to show an arrangement like $$ggbggbggbggbggbggbgb$$ will also suffice

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    $\begingroup$ Well, you can show that the pattern $bgg$ can't exist either. $\endgroup$ – harvey Aug 4 '16 at 5:02
  • $\begingroup$ By "some" do you mean a non-zero quantity? Because as it stands, I can't see why there can't be 20 boys. $\endgroup$ – Mr. Chip Aug 4 '16 at 5:22
  • $\begingroup$ Yes 👽 i needed to edit my initial post that the sequence of ggb's ** IS NOT ** permissible ( my bad but understood why )( hence a ' typo') $\endgroup$ – Randin Aug 4 '16 at 7:18
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Three possibilities are to have no boys, all boys, or 10 boys alternating with girls.

Number the seats 1-20 clockwise. If the person in seat 1 is a boy, then person 2 wears blue, so person 2 can't wear red, so person 3 can't be a girl. In general, if seat $n$ has a boy, then seat $n+2$ has a boy as well. Similarly, if $n$ has a girl, then $n-2$ has a girl as well. Iterating this argument shows that all odd numbered seats have the same gender, and same for even ones, proving those first three patterns are the only ones possible.

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The two conditions can be modelled with sticks. "Right of boy (M) is blue (B) T-shirt" and "left of girl (F) is red (R) T-shirt" are as below:

--B-R--
|/| |\|
M-----F

Now put in a boy:

--B----
|/| | |
M-1----

If I am to extend this configuration, I may put at (1) a girl:

R-B----
|X| | | (A)
M-F-2--

or at (1) a second boy:

--B-B--
|/|/| | (B)
M-M-2--

In (B), I cannot place a girl at (2) because it contradicts the second boy's T-shirt colour. Therefore I must place a third boy:

--B-B-B
|/|/|/|
M-M-M--

and I am forced to continue this pattern right around the circle, whereupon I have 20 boys, all wearing blue shirts.

In (A) I must place a boy at (2), since placing a girl there would again introduce a contradiction:

R-B---B
|X| |/| (C)
M-F-M-3

and if I now place a boy at (3) I am forced around the circle like (B):

|   20 seats    |  
R-B---B-B...B-B-B-R-B
|X| |/|/|...|/|/| |X|
M-F-M-M-M...M-M-4-M-F

But then (4) cannot be assigned boy or girl – another contradiction. Thus I must place a girl at (3) in (C):

R-B-R-B
|X| |X|
M-F-M-F

Inductively, I have the same conditions as (A), but with two fewer seats. Hence I must continue the pattern, and I end up with 10 boys with red shirts and 10 girls with blue shirts alternating.

This, of course, assumes you have at least one boy; you could also have 20 girls with red shirts around the table. (This is just the 20-boys solution with "boy" replaced by "girl" and "blue" by "red".)

The number of boys can thus be only 0, 10 or 20.

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$1$: If there exists a boy with a blue t-shirt, then the right neighbour of this boy must be a boy with blue t-shirt again ( it cannot be a girl because the left neighbour has a blue t-shirt). Continuing this process, you will obtain 20 boys.

$2$: If there exists a girl with a red t-shirt, then the left neighbor of this girl must be a girl with red T-shirt again (it cannot be a boy because the right neighbour has a red t-shirt). Continuing this process, you will obtain 20 girls.

$3$: Let there exist at least one boy with a red T-shirt (in this instance, we know that there does not exist any boy with a blue t-shirt; see $1$). The neighbour to the right of this boy must be a girl with a blue t-shirt and the neighbour to the right of this girl must be a boy with a red t-shirt. Continuing this process, you will obtain ten boys with red t-shirts and ten girls with blue t-shirts.

So the number of boys can be $0$ or $10$ or $20$

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