In how many different orders could I eat the fruit?

Hey so basically I have a scholarship coming up and one of the questions in the sample paper is this:

"In my packed lunch I have 3 strawberries, 2 cherries and one apple. In how many different orders could I eat the fruit?(one order is strawberry,cherry,apple,cherry,strawberry,strawberry)"

I don't want just the answer but a way to work out others like it

thanks in advance and I'm so grateful!!

Kitch said:

"In my packed lunch I have 3 strawberries, 2 cherries and one apple. In how many different orders could I eat the fruit?(one order is strawberry,cherry,apple,cherry,strawberry,strawberry)"

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Are you given any instructions regarding how to "view" the fruits? For instance, should the cherries be viewed as "indistinguishable", so they're interchangeable, or should they be viewed as "distinguishable", so there is a difference in which you pick first? In terms of setting up this exercise, in other words, should the cherries be denoted as "C" and "C" (so they're interchangeable) or as "C₁" and "C₂" (so they're different)?

Thank you!

Reply

No that's the whole question but I guess that it means different cherries
They don't have to stick as a group

Kitch said:

"In my packed lunch I have 3 strawberries, 2 cherries and one apple. In how many different orders could I eat the fruit?(one order is strawberry,cherry,apple,cherry,strawberry,strawberry)"

Click to expand...

Having seen this sort of question many times, I can almost assure you that the fruit are considered identical within their class. It is asking 'how many orders in terms of class are possible'.

That no different from asking, "how many arrangements are possible of the string \(\displaystyle ssscca~?\)

The answer is \(\displaystyle \dfrac{6!}{3!\cdot 2!}\).

Thank you but can u plz explain?
Also on the 6! / 3!.2!
what does the point mean

Kitch said:

Also on the 6! / 3!.2!
what does the point mean

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It means multiplication: \(\displaystyle 3!\cdot 2!=(3!)(2!)\).

We we have strings with multiple repeated characters we must account for them.

The number of ways to rearrange the word \(\displaystyle MISSISSIPPI\) is \(\displaystyle \dfrac{11!}{(4!)^2(2!)}\), because there are four I's, four S's, and two P's.

Kitch said:

Hey so basically I have a scholarship coming up and one of the questions in the sample paper is this:

"In my packed lunch I have 3 strawberries, 2 cherries and one apple. In how many different orders could I eat the fruit?(one order is strawberry,cherry,apple,cherry,strawberry,strawberry)"

I don't want just the answer but a way to work out others like it

thanks in advance and I'm so grateful!!

Click to expand...

pka said:

Having seen this sort of question many times, I can almost assure you that the fruit are considered identical within their class. It is asking 'how many orders in terms of class are possible'.

That no different from asking, "how many arrangements are possible of the string [FONT=MathJax_Math]s[FONT=MathJax_Math]s[/FONT][FONT=MathJax_Math]s[/FONT][FONT=MathJax_Math]c[/FONT][FONT=MathJax_Math]c[/FONT][FONT=MathJax_Math]a[/FONT][FONT=MathJax_Main]?
[/FONT]
The answer is \(\displaystyle \dfrac{6!}{3!\cdot 2!}\)[/FONT]

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The centered dot is one way to indicate multiplication.
You have 6 objects to arrange. 6 choices for what is first, 5 choices for 2nd, . . .
6 * 5 * 4 * 3 * 3 * 1 = 6!
is the numerator of the result

BUT not all of those permutations are unique, because some of the objects are identical. The two cherries can always be interchanged, so you have to divide by 2. [Note: for symmetry and generality, we write 2! instead of 2.] Likewise if the three strawberries are rearranged, you can't tell the difference. Since 3 objects can be arranged 3*2*1 ways, you have to divide by 3!.

Does that explain where the answer came from? The general result for combinations of N objects including some classes of identical objects is

\(\displaystyle \dfrac{N!}{n_1!\ \cdot\ n_2!\ \cdot\ \cdot\ \cdot}\)

I am not giving you a different answer but explaining the thought process behind the answer. Let's say you put little stickers on the pieces of fruits from a through f. So you could eat a first, or b first, or c first. There are 6 choices on what you eat first. That leaves 5 choices on what you eat second. So there are 6 * 5 = 30 choices on what you eat first and second. There are 4 choices on what you eat third. So there are
6 * 5 * 4 = 120 choices on what to eat first, second, and third. So ultimately there are 6 * 5 * 4 * 3 * 2 * 1 = 720. This kind of product comes up so often there is a special symbol for it 6! = 6 * 5 * 4 * 3 * 2 * 1.

With me so far.

Now what pka did was to recognize that there are no stickers. So let's consider any arrangement. If there were stickers a and b on the cherries and c, d, and e on the strawberries, one possible sequence would be fabcde and another would be fbacde. If there were no stickers on the cherries, you would not be able to tell those two sequences apart. That means that 6! is an overcount. If there were no stickers on the strawberries, you would not be able to tell these six sequences apart
fabcde, fabced, fabdce, fabdec, fabecd, and fabedc. Again that means that 6! is an overcount. So the answer is

6! / (2 * 6) = 6! / (2! * 3!)

This is all studied under the name of permutations and combinations.