-- He must have at least one of each color in the case, so the first 3 of the 5 marbles in the case are blue-green-black.

Now the rest of the collection consists of

4 blue
4 green
      2 black

and there's space for 2 more marbles in the case.

So the question really asks: "In how many ways can 2 marbles
be selected from 4 blue ones, 4 green ones, and 2 black ones ?"

-- Well, there are 10 marbles all together.
So the first one chosen can be any one of the 10,
   and for each of those,
the second one can be any one of the remaining 9 .

Total number of ways to pick 2 out of the 10 = (10 x 9) = 90 ways.

-- BUT ... there are not nearly that many different combinations
to wind up with in the case.

The first of the two picks can be any one of the 3 colors,
   and for each of those,
the second pick can also be any one of the 3 colors.

So there are actually only 9 distinguishable ways (ways that
you can tell apart) to pick the last two marbles.

6 0

3 years ago

At one baseball stadium, there are 4 kinds of seats .what fraction of the total seats are premium

bija089 [108]

Assuming that each kind of seat is unnique and different from each other
( assume 4 kinds, premium, gold, silver, bronze, or premium, not premium, not premium, not premium)

so fraciton is parts out of whole so

1 out of 4 toal parts is 1/4

7 0

3 years ago

What's The Answer??????????????????

Yuri [45]