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I'm working on a problem as follows. I think my answers are correct, as they logically make sense when I work through them (d is a little shaky). Given that these are rarely as simple as 2+2 = 4, I wanted to run these by someone for a second opinion ;P

Question: A, B, C and D are four of fifty people who attend a prize drawing. Each of the 50 people has a single ticket in drawing. There is one grand prize (new bike) and 3 identical secondary prizes (helmets. How many ways to award prizes with:

$A)$ No additional restrictions: $50 * C(49,3) = 921,200$
50 possible winners for the bike, then 49 choose 3 for the helmets?

$B)$ If A wins a secondary prize: $50 * 1 * C(48,2) = 56,400$
50 possible winners for bike, anna definitely wins (1) then 48 choose 2 for helmets?

$C)$ Either A, B, C or D wins the grand prize: $C(49,3) * 4 = 73,696$

If one of them wins bike, it's $1 * C(49,3)$ So the number of ways for either A or B or C or D to win the bike is that number multiplied by 4?

$D)$ A and B both with prizes (of some sort) while C and D do not win prizes.
I don't actually know how to do this. I've tried breaking it down in balls and bins (which doesn't seem to work) and I've also thought about it as a deck of cards. What would a similar question be using either of those two models to help change my way of thinking?

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A wins major prize, d wins secondary: ${46 \choose 2}$ Then the opposite case where D wins the major prize and A a secondary: ${46 \choose 2}$ (again)

So the answer is $$2{46 \choose 2} = 46 \times 45 = 2070 $$

Note: we are working with disjoint cases (their intersection is null, that means that they are mutually exclusive cases) so we can add them up.

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  • $\begingroup$ You are choosing 2 from 46 to remove B & C from the running? $\endgroup$ – Podo Dec 1 '17 at 1:57
  • $\begingroup$ Exactly, I'm just ignoring those 2 $\endgroup$ – Francisco José Letterio Dec 1 '17 at 1:58
  • $\begingroup$ So the way I solved A B And C are similar to how you solved D, I'll assume I approached those correctly. Thanks for the help mate. $\endgroup$ – Podo Dec 1 '17 at 2:10
  • $\begingroup$ Yes, those were correct (at least I cant find any fault) $\endgroup$ – Francisco José Letterio Dec 1 '17 at 2:12

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