If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# Generalizing with binomial coefficients (bit advanced)

Conceptual understanding of where the formula for binomial coefficients come from. Created by Sal Khan.

## Want to join the conversation?

• I can never make sense of where the (n-(k-1)) bit comes from... •   You can think of it as a way to limit the number of terms in

n . (n - 1) . (n - 2) . (n - 3)...

We want K terms here - one for each heads that will appear in our coin flips. If you look at the digits in those terms, they're nicely numbered: 1, 2, 3...

Since we want K terms, we nearly want a set of terms that goes 1, 2, 3, all the way up to K. BUT! Our first term is just n on its own. This means the last term we want isn't (n - K), but the one before it in the sequence: (n - (K - 1)).
• How do i apply this method if it's not fair coins? if there is a larger percentage to get heads than tails?
Best of regards, • I think the easiest way is just to add up all probabilities of exact arragments.
for example, we have p% of probability of getting heads.
therefore probability of getting exactly n heads in m flips:
(p%)^n * (1-p%)^(m-n) * ( mCn )
mCn is binomial coefficients.
(1-p%) is probablity of getting tails.
• Is there a formula for calculating (or rules for addin up) probability of pulling AT LEAST some amount (k) of similar items from a bag of (n) mixed items, when you get to take out x amount of items from the bag?

E.g. a bag has 8 colored beads, 4 of them are blue beads. What's the probability of getting AT LEAST 3 blue beads (i.e. 3 or 4 blue beads) when you can take out 4 beads.

I've tried figuring this out but could use some help. • Why does he say the "last" video when the last video was about the mega millions jackpot???? • At he says that you can use this formula for finding the number of ways to stick 5 things in 2 chairs without differentiating between those two things. What does he mean when he says differentiate? • He means essentially that we don't care if Object #1 and Object #2 are different, we won't be thinking of them as different.

So say you have 5 people that you need to put into two chairs. One chair is on the left, the other is on the right. Jack and Jill are two different people, but we don't care if Jack is on the left and Jill on the right, or if Jill is on the left and Jack on the right. All we care is that we have Jack and Jill, instead of Jack and Sally, or Bob and Charles, etc.
• Why does Sal write × (times) symbols as . (decimal points)? • The dot operator is a very commonly used symbol for multiplication. In my experience, the dot (` · `) notation is much more common than the times sign (` × `). The × sign can sometimes get confused for the letter X, which we would really like to avoid doing. When dealing with multiple variables, it's also fairly common to not have a multiplication sign at all - multiple letters next to each other just implies multiplying the together. The · or × can be inserted when substituting numbers in for variables, or you can just use parentheses.

Similarly for division, using the solidus (` / `) is much more common than the obelus (` ÷ `) for "inline" fractions (that is: a faction written like a line of text, instead of being vertically aligned). For example:
`(a-b) / c` instead of `(a-b) ÷ c`
• I don't understand about marking the heads A, B etc. You either flip one coin 5 times in a row or flip 5 coin simultaneously, what is he talking about with the A, B, C etc.
It's like you're saying you choose coin A, then choose where to insert that coin in the 5 possibilities/spaces (like space 1, 2 etc.). • It doesn't matter if you flip 1 coin 5 times of 5 coins simultaneously, what does makes a difference is the "bucket" you assign to each coin.

Think of it this way: You have 5 "slots" to fill with coin flip results. How you fill then is left to you, if you flip just 1 coin you say: I'm flipping the coin and the results goes into "slot 1". Now I'm doing it again and the result goes into slot 2, and co... If OTOH you would have 5 coins you would say: coin 1 flip results goes into slot 1, coin 2 into slot 2. Now flip them all and lets see the results.
• I'm a bit confused here, what happens when k = n, like in 5 heads in 5 flips? It seems the denominator goes to zero :S What am I getting wrong?

Edit: Aha I think I got it, if k = n only k! should be left in the denominator, no need for the (n-k)! term, right?   