How many ways can I choose 5 items from 10?

Question

I'm learning about factorials and it looks like choices (how many ways you can choose something) is related to factorial.

Does anyone know how many ways can I choose 5 items from 10? (For example, 5 balls from 10.)

I think it's $\frac{5!}{10!}$. Am I right?

Edit: I mean $\frac{10!}{5!}$

Answer 1

There are $\dfrac{n!}{k!(n-k)!}$ ways to choose $k$ things from $n$ things.

Answer 2

Here is a copy of an answer I gave to a similar post (I'll edit it to fit your question a bit better):

The formula for permutations is $\frac{n!}{(n-k)!}$
The formula for combinations is $\frac{n!}{k!(n-k)!}$

(Here I assume you understand the permutation formula a bit). To derive the second from the first let's make an example... say we are finding the combinations of $3$ numbers from a set of $4$, say $1,2,3,$ and $4$. If we use the permutation formula we get our answer as $24$, but we can see we over-counted... for example, $1,2,3$ and $1,3,2$ were counted separately. Since we are taking sets of size $3$ and there are $3!$ ways to arrange a set of size $3$, we have over-counted by a factor of $3!$. Generalizing this, if we use the permutation formula to get combinations we will always over-count by $k!$, where $k$ is the size of the set we take from the whole, and hence the formula for combinations above (we say order does not matter with combinations because of that division... every set we take is unique no matter how we arrange it. This is not true for permutations).

EDIT: Since you are new to this, we get the permutations formula by first choosing one from a set of $n$, then $(n-1)$, then $(n-2)$, and we do this until we get $(n-k+1)$, yielding $P(n,k)=n(n-1)(n-2)(n-3)\ldots(n-k+1)$ (The last case avoids $0$ when $n=k$). We then multiply by (n-k)! to get $$P(n,k)=\frac{n(n-1)(n-2)(n-3)\ldots(n-k+1)(n-r)!}{(n-r)!}$$ $$= \frac{[n(n-1)(n-2)\ldots(n-k+1)]\cdot[(n-k)(n-k-1)\ldots(3)(2)(1)]}{((n-r)!}$$ $$= \frac{n!}{(n-r)!}$$ (The last step requires an observation about the descending nature of the numerator... I leave it as a math puzzle to you!)

Answer 3

Well, it's $\frac{10!}{5!}$, if you're counting each order you select balls as a different "way." For each of those $\frac{10!}{5!}$ sets of $5$ balls, there's $5!$ ways to order it, so one way to see the correct answer is to divide by the number of same re-orderings, and get $\frac{10!}{5!5!}$, which is correct.

(I always found this explanation hand-wavey, but it does show the difference between what you've come up with so far, and the correct answer.)

Answer 4

I will encourage you to learn combinatorics basics.

Let me explain the answer. let $X=\{x_1,x_2,x_3,x_4,x_5,x_6,x_7,x_8,x_9,x_{10}\}$ be the 10 different items you want to choose $5$ items from. This will be the total no. of subsets of $X$ with size $5$. Let $A$ denotes the set of all possible subsets with size $5$.
Then there is natural bijection from $A$ to the set of n-bit sequences defined as as:
$b_i=1$ if $i^{th}$ element is chosen
$b_i=0$ otherwise.
Since we have to choose 5 items, then let's randomly denote the above sequence as $(0,0,1,0,1,0,0,1,1,1)$ . Let's say $B$ be set of all possible sequences.
Then the mapping $f:A\rightarrow B$ defined as $a \mapsto b$ $s.t.$ $a$ has 5 items with the corresponding $b_i=1$.
Since $f$ is a bijection, therefore:
We are interested in $$|A|=|B|=\frac{10!}{5!.5!}=\binom{10}{5}$$

Answer 5

Consider 10 items (a,b,c,d,e,f,g,h,i,j) There are 10 different options of choosing first number, say a if chosen. This leaves 9 different choices for second, say c is chosen. This leaves 8 different choices for third, say d is chosen. This leaves 7 different choices for fourth, say f is chosen. This leaves 6 different choices for fifth, say h is chosen.

So (a,c,d,f,h) has been selected but total possible was 10x9x8x7x6 which using factorial notation can be written as 10!/5!

Now if the order is not important and only the combination is needed, then (a,c,d,f,h) is the same as (d,a,f,h,c). This means that we need to divide the answer by all the possible patterns (permutations) for five letters. By similar argument the possible number of options for choosing the first is 5, the second is 4 and so on giving the answer to be 5x4x3x2x1 which can be written as 5! So the number of combinations of 5 items from 10 is 10!/(5!5!).

Note 5! Appears twice in our denominator because it was the number used in original question. If we had been choosing 4 from 10 the answer would have been 10!/(4!6!)

General equations are number of permutations (i.e. patterns) of r objects from n objects, all objects being different is n!/r!

General equations are number of combinations of r objects from n objects, all objects being different is n!/((r!(n-r)!)