0

A certain number of boys and girls can be seated in a row such that no two girls are together in $1440$ ways. If one more boy joins them, the number of ways in which they can be seated in a row such that no two girls are together increases:
a) 4 fold

b) 6 fold

c) 8 fold

d) 10-fold

I am not getting how to solve this problem though I have an intuition that there's some usage of $6!\times 2 = 1440$

Reason: Let's assume $x$ boys and $y$ girls, $1440 = x! \times y!$

Archer
  • 6,591

3 Answers3

3

Suppose there are $b$ boys and $g$ girls. The boys can be arranged in $b!$ ways. In each such arrangement, there are $b + 1$ spaces in which we can place the girls, $b - 1$ between successive boys and two at the ends of the row. To separate the girls, we must choose $g$ of these $b + 1$ spaces in which to place a girl (assuming that is possible). The girls can be arranged in the selected spaces in $g!$ ways. Hence, the number of ways $b$ boys and $g$ girls can be arranged in a row so that no two of the girls are adjacent is $$b!\binom{b + 1}{g}g!$$ You need to find $b$ and $g$ so that this quantity equals $1440$, then modify the above formula for $b + 1$ boys and $g$ girls, from which you can determine the factor.

N. F. Taussig
  • 79,074
  • It is possible. Clearly, $b < 7$ since $7! = 5040 > 1440$. Start with $b = 6$. If it does not work, try smaller values for $b$. – N. F. Taussig Jan 04 '18 at 18:55
1

Suppose there are $x$ boys and $y$ girls.

If no two girls are together, then $x+1\ge y$

That is, if there are is one more girl than boys, we can put a girl on each end, and proceed girl, boy, girl, boy. If there are fewer boys, there is no arrangement with no two girls together.

If we have more than the minimum number of boys, we have ${x+1\choose y}$ ways we can arrange the "surplus boys."

$x!y!\frac {(x+1)!}{y!(x-y+1)!} = \frac {x!(x+1)!}{(x-y+1)!} = 1440$

Since $1440$ is divisible by $5, x+1 \ge 5$ or we don't get a factor of $5$ into the numerator.

Quickly we find that $(x,y) = (4,3)$ is the only pair that gives the correct result.

adding one more boy.

$\frac {(x+1)!(x+2)!}{(x-y+2)!} = \frac {x!(x+1)!}{(x-y+1)!}\frac {(x+1)(x+2)}{x-y+2} = 1440\cdot {5\cdot 6}{3} = 1440\cdot 10$

Doug M
  • 58,694
0

Your reasoning is slightly wrong; $x!\times y!$ is the number of ways of seating $x$ boys and $y$ girls, but you're missing the factor due to the number of arrangements of (unlabeled) girls among the (equally unlabeled) boys. Instead, I think you should treat this as a (slightly modified) Stars And Bars problem, and consider the problem from that perspective.

Steven Stadnicki
  • 53,147
  • Please tell me how I should treat it as a stars and bars problem – Archer Jan 04 '18 at 18:21
  • @Abcd See, for instance, https://math.stackexchange.com/questions/483071/ - if you look in the sidebar you'll see many related questions, some of which cover exactly this case. – Steven Stadnicki Jan 04 '18 at 18:26