In how many ways can 10 different books be divided among a, b and c so that 5 given to A, to 3 to B and 2 to C.
Well, if we're dividing the books among A, B, and C, it's like organizing a library with three librarians. Let's see how this plays out in a humorous way:
Librarian A, being the most bookwormy of the bunch, insists on taking 5 books. Maybe they're all murder mysteries or romance novels – who knows?
Librarian B, slightly envious of Librarian A's selection, decides they can manage with just 3 books. Maybe they prefer fantasy or self-help books – to each their own!
Finally, Librarian C is content with only 2 books. They're probably busy enough keeping the library in order and don't have much time for reading.
Now, let's think about the process. We could start by picking 5 books for Librarian A out of the 10 available. That would leave us with 5 books remaining. Then we pick 3 books out of those for Librarian B, which would leave us with 2 books left. Finally, we give those last 2 books to Librarian C.
So, the number of ways to divide the books would be the product of the choices for each librarian: 10C5 * 5C3 * 2C2.
Simplifying that, we get: (10! / (5! * (10-5)!) * (5! / (3! * (5-3)!) * (2! / (2! * (2-2)!).
Let me just calculate that for you... *calculating noises*
It turns out there are 2520 different ways to divide the books among A, B, and C according to your requirements.
But hey, don't take my word for it. Feel free to double-check my calculations and remember to ask Librarian A for some book recommendations – they've got quite the collection!
To find the number of ways to divide 10 different books among A, B, and C, you can use the concept of combinations.
Step 1: Distribute the 5 books to A. There is only one way to do this.
Step 2: Distribute the 3 books to B. There is only one way to do this.
Step 3: Distribute the 2 books to C. There is only one way to do this.
Therefore, the total number of ways to divide the books among A, B, and C is 1 * 1 * 1 = 1.
So, there is only one way to divide the 10 different books among A, B, and C according to the given conditions.
To find the number of ways the 10 different books can be divided among A, B, and C with specific quantities, we can use the concept of combinations in combinatorial mathematics.
First, we need to allocate the books as mentioned: 5 books to A, 3 books to B, and 2 books to C. Since the books are different, we are considering the order in which they are given to A, B, and C.
To distribute the books to A, we choose 5 books out of the 10 available. This can be expressed as "10 choose 5" or written as C(10, 5). The formula for combinations is given by C(n, r) = n! / (r! * (n - r)!), where n is the total number of objects, and r is the number of objects selected.
Using this formula, C(10, 5) = 10! / (5! * (10 - 5)!) = 10! / (5! * 5!) = (10 * 9 * 8 * 7 * 6) / (5 * 4 * 3 * 2 * 1) = 252.
Next, for books given to B, we choose 3 books out of the remaining 5. This can be expressed as C(5, 3) = 5! / (3! * (5 - 3)!) = 5! / (3! * 2!) = (5 * 4) / (2 * 1) = 10.
Finally, for books given to C, we don't have any choice as there are only 2 books remaining.
To find the total number of ways, we multiply the number of ways for each person: 252 * 10 * 1 = 2,520.
Therefore, there are 2,520 ways to distribute 10 different books among A, B, and C, with A receiving 5 books, B receiving 3 books, and C receiving 2 books.
I don't suppose that the order of assigning the books matters so we have combinations.
"a" can be assigned books in C(10,5) ways, leaving 5 books for "b" who needs 3, that would be C(5,3), leaving the last 2 books to "c",
number of ways = C(10,5) x C(5,3) x C(2,2)
= 252*10*1
= 2520