Question

An urn contains 9 white balls and 10 red balls. Three balls are selected. In how many ways can the 3 balls be drawn from the total of 19 balls:
a) If 2 balls are white and 1 is red?
b) If all 3 balls are white?
c) If all 3 balls are red?

Answer 1

same as the last one

Answer 2

two white, one red
\[\binom{9}{2}\times \binom{10}{1}\] which is really just
\[\frac{9\times 8}{2}\times 10\]

Answer 3

3 white
\[\binom{9}{3}\]
3 red \[\binom{10}{3}\]

Answer 4

so do you always just subtract 1 for the like 9x8 thing (like where did you get 8?)

Answer 5

lets compute 9 choose 2 by the counting principle

Answer 6

9 choices for the first ball
then there are 8 left, so 8 choices for the second ball
by the counting principle, the number of ways to do that is \(9\times 8\)
then you have to divide by 2 because the above method counts \((a,b)\) differently from \((b,a)\)

Answer 7

similarly
\[\binom{9}{3}=\frac{9\times 8\times 7}{3\times 2}\]

Answer 8

ok! so is it a) 360
b) 84
c) 120

Answer 9

\[\binom{12}{4}=\frac{12\times 11\times 10\times 9}{4\times 3\times 2}\] sometimes written as
\[\frac{12\times 11\times 10\times 9}{4!}\]