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?


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.