permutations problem - student may not win all 23 prizes

• May 8th 2011, 04:09 AM
angypangy
permutations problem - student may not win all 23 prizes
Problem is:
3 prizes, one for English, one for French and one for Spanish can be awarded in a class of 20 students. Find the number of different ways in which the three prizes can be awarded if:
(i) no student may win > 1 prize.
(ii) no student may win all 3 prizes.

(i) I can do ok - 20 * 19 * 18 = 6840. ie 20 students can win E, leaving 19 to win F, etc.

But I am struggling with (ii). If the question asked was student CAN win 1,2 or 3 prizes then poss ways would be 20*20*20=8000. Is that correct.

My first thought was that way to do this is 20*20*19 - but that is incorrect. Correct way to solve is apparently 20*20*20 - 20. But why?

Angus
• May 8th 2011, 04:21 AM
Plato
Quote:

Originally Posted by angypangy
3 prizes, one for English, one for French and one for Spanish can be awarded in a class of 20 students. Find the number of different ways in which the three prizes can be awarded if:
(ii) no student may win all 3 prizes.
My first thought was that way to do this is 20*20*19 - but that is incorrect. Correct way to solve is apparently 20*20*20 - 20. But why?

There are \$\displaystyle 20^3\$ functions from a set of three to a set of twenty. But there are 20 ways that all three go to the same image (in this case a student). So subtract.
\$\displaystyle 20^3-20\$.
• May 8th 2011, 04:26 AM
angypangy
Sorry to be a pain but how do you calculate the 20. you say 20 ways that all three go to the same image? How are you getting the 20 to subtract?
• May 8th 2011, 04:39 AM
angypangy
Its ok I have got it now.
• May 8th 2011, 04:39 AM
Plato
Quote:

Originally Posted by angypangy
Sorry to be a pain but how do you calculate the 20. you say 20 ways that all three go to the same image? How are you getting the 20 to subtract?

How many ways can you give all three prizes to the same student?
Those ane the cases you want to exclude (no student gets all three).