combinations with fruits (1 Viewer)

[facebooknerd123]

there are 4 oranges, 3 bananas and 2 apples, what is the amount of ways to choose at least one fruit of each if you pick five in total?

someone pls explain why it is not 4C1 multiplied by 3C1 multiplied by 2C1 multiplied by 6C2 ?
because pick 1 orange, 1 banana, 1 apple and pick everything else (2 left over)
It equals to like 360 wtf? Even 9C5 (total) doesn't exceed 360

 

[porcupinetree]

Each fruit is not unique within its own type. i.e., all the bananas are identical.

Solution: we already have confirmed that we'll have 1 of each fruit. Hence we just need 2 more. There are 5 ways to do this total, if you count them up.

 

[facebooknerd123]

Each fruit is not unique within its own type. i.e., all the bananas are identical.

Solution: we already have confirmed that we'll have 1 of each fruit. Hence we just need 2 more. There are 5 ways to do this total, if you count them up.

the answer is 98. I did this by cases which was easier. like i did total - [0 orange, 4 banana, 0 apple + ... ] But why my way not work? i onyl choose 1 banana , ... , 1 apple and i did not arrange anything

 

[porcupinetree]

the answer is 98. I did this by cases which was easier. like i did total - [0 orange, 4 banana, 0 apple + ... ] But why my way not work? i onyl choose 1 banana , ... , 1 apple and i did not arrange anything

Well it depends upon whether the fruits are considered to be unique within their type (i.e. can you tell the difference between banana 1 and banana 2?)

 

[facebooknerd123]

probably not. so what would be the working out for it?

 

[braintic]

probably not. so what would be the working out for it?

If each type of fruit is considered to contain identical pieces of fruit, you simply list out the cases:

3 1 1
2 2 1
2 1 2
1 3 1
1 2 2

5 cases

If the fruits are not considered identical, you do a calculation for each of those cases:

4 × 3 × 2
6 × 3 × 2
6 × 3 × 1
4 × 1 × 2
4 × 3 × 1

Adding those up gives 98

 

[facebooknerd123]

thanks, how come my way did not work? where I did 4C1 * 3C1 ...

 

[braintic]

thanks, how come my way did not work? where I did 4C1 * 3C1 ...

Because you are double counting.

If the fruits are labelled O1, O2, O3, O4, B1, B2, B3, A1, A2:

If you choose O1, B1, A1 as your initial 3 pieces of fruit, then choose O2, B2 for your last two pieces,
that is exactly the same as choosing O1, B2, A1 as your initial 3 pieces of fruit, then choosing O2, B1 for your last two pieces.

 

[facebooknerd123]

that makes a lot of sense. But how can I eliminate the double counting? What do I subtract with?

 

[braintic]

that makes a lot of sense. But how can I eliminate the double counting? What do I subtract with?

The factor you are overcounting by depends on the number of each type of fruit you end up with.
ie. OOOAB and OOAAB will be overcounted by different amounts.
As you still have to take multiple cases, it is easier to use the other method.