permutations/combinations problem

In how many ways can four boxes of chocolate be chosen from a store with 10 different selections if at least two of the boxes are of the same type?

okay for this one, i counted cases
1) all 4 the same: 4C4
2) 3 same, 1 different : 4C3 x 7C1
3) 2 same : 4C2 x 8C2
and then i just add these but apparently the answer is 505 so i'm confused.

jijie14 said:

In how many ways can four boxes of chocolate be chosen from a store with 10 different selections if at least two of the boxes are of the same type?

okay for this one, i counted cases
1) all 4 the same: 4C4

Click to expand...

There are ten different assortments. So shouldn't "all four boxes are the same assortment" have ten ways to be chosen?

jijie14 said:

2) 3 same, 1 different : 4C3 x 7C1

Click to expand...

In how many ways can you choose the first assortment? How many options are there for that? In how many ways can you choose the second assortment? How many options are there for that? :wink:

Hello, jijie14!

Your counting is faulty.

In how many ways can 4 boxes of chocolate be chosen from a store with 10 different selections
if at least two of the boxes are of the same type?

Click to expand...

Reminder: There are 10 types of chocolates.
. . . . . . . . We will select 4 of them.


4 of one type:
. . \(\displaystyle 10\) choices.

3 of one type, 1 of another:
. . \(\displaystyle 10\cdot9 \,=\,90\) choices.

2 of one type, 2 of another:
. . \(\displaystyle {10\choose2} \,=\,45\) choices.

2 of one type, 1 of another, 1 of another:
. . \(\displaystyle 10\cdot{9\choose2} \,=\,360\) choices.


Therefore, there are: .\(\displaystyle 10 + 90 + 45 + 360 \:=\:505\) ways.