Don't Separate Us

Probability Level 2

"DeeDah", as he is playfully called by all his grandchildren, likes all of his extended family. DeeDah goes for his evening walks, always taking exactly 4 of his grandchildren at a time. That is the maximum craziness he can handle at a time!!

Now, DeeDah never takes the same group. However, another concern is that Macy and Ryan refuse to be separated, so that either they will go together or not at all.

If Macy and Ryan remove their inseparability clause, then DeeDah would have had 330 more options with him in the evening.

How many crazy yet indispensable grandchildren does DeeDah have?

13 11 15 9

1 solution

Brian Charlesworth
Feb 24, 2018

Suppose Deedah has $n$ grandchildren. With the inseparability clause in effect, Deedah could take $4$ of the remaining $n - 2$ grandchildren in $\dbinom{n - 2}{4}$ ways, or take Macy and Ryan along with $2$ of the remaining $n - 2$ grandchildren in $\dbinom{n - 2}{2}$ ways. If Macy and Ryan remove their inseparability clause then Deedah would have $\dbinom{n}{4}$ options. So the equation we need to solve is

$\dbinom{n}{4} - \left(\dbinom{n - 2}{4} + \dbinom{n - 2}{2}\right) = 330$

$\Longrightarrow \dfrac{n(n - 1)(n - 2)(n - 3)}{24} - \dfrac{(n - 2)(n - 3)(n - 4)(n - 5)}{24} - \dfrac{(n - 2)(n - 3)}{2} = 330$

$\Longrightarrow \dfrac{(n - 2)(n - 3)}{24}(n(n - 1) - (n - 4)(n - 5) - 12) = 330$

$\Longrightarrow \dfrac{(n - 2)(n - 3)}{24}(8n - 32) = 330 \Longrightarrow (n - 2)(n - 3)(n - 4) = 990$ .

By observation, as $990 = 9 \times 10 \times 11$ is the product of 3 consecutive integers we can conclude that $n - 2 = 11 \Longrightarrow n = \boxed{13}$ .

(Note that as $(n - 2)(n - 3)(n - 4) - 990 = (n - 13)(n^{2} + 4n + 78)$ the other 2 possible values for $n$ are complex.)

Don't Separate Us

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