Printable View
A jar contains ten blue balls, ten red ones, ten white ones, and ten green ones. Each set of ten balls of the same colour is numbered from 1 to 10. How many different selections of four balls are possible if exactly three of the balls bear the same numbers?
The answer's supposed to be 1440.
Can someone show me the steps of solving this problem? Thanks!
CASE (i)Quote:
Originally Posted by wabbt
suppose the three balls bearing the same number bear the number 1.
there are four subcases:
1) those three balls are B,R,G
2) those three balls are B,R,W
3) those three balls are R,G,W
4) those three balls are B,G,W
for subcase 1) there are 36 possible choices. (why?)
So for CASE(i) there are 4*36 choices.
Hence the total number of choices are 4*10*36.
There ten ways to pick the number that occurs three times.Quote:
Originally Posted by wabbt
The there are thirty-six ways we can pick the odd-man out.
There are four ways each of those selections can be made.
How come it's 36 instead of 37? Since we only need 3 balls of the same color and there would be 37 left to choose from?
Once we have the number to be used three times in effect we have ruled out four balls because we cannot the forth ball of that number. That leaves thirty-six to choose.Quote:
Originally Posted by wabbt
Hello, wabbt!
Quote:
A jar contains ten blue, ten red, ten white, and ten green balls.
Each set of ten balls of the same colour is numbered from 1 to 10.
How many different selections of four balls are possible
if exactly three of the balls bear the same numbers?
The answer's supposed to be 1440.
Consider this analogous problem . . .
Remove the face cards from a standard deck of cards.
You 40 cards, from Ace to Ten in four suits.
You draw four cards.
In how many ways can you get Three-of-a-Kind?
The are 10 choices for the value of the triple.
There areways to get the triple.
There are 36 choices for the fourth card.
. . (We do not want Four-of-a-Kind.)
Therefore, there are: .ways.