How To Understand Combinations Using Multiplication

Greta article. One comment. Under ‘Where Next’ you wrote ’ This is what the binomial theorem does. We’ll cover more later…’ Where does ‘more later’ refer to?

Thank you for your explanation.

Do you mind reading this post to see if it make sense? I’m not a person that is naturally comfortable with anything connected with Stats (I’m trying hard though).

On your comment about the binomial expansion : choose a or b 10 times?

Is it that you have an a OR b event and you are repeating this event 10 times. This means you are picking an outcome from each experiment. Then for each term you are counting the number of possible strings of experiments that lead to having say choosen three a’s.

Choosing 3 a’s forces you to have choosen 7 b’s as you need to choose one in each experiment.

Also it cannot be anything else as with our other hat one we are still talking about multiplication and the dimensions have to be consistant. Multiplication can be though of as finding the area/volume/measure of the “rectangliod” of the number of dimensions the same as the number of numbers you are multiplying. Addition can only be performed between things that agree in dimension (you can’t add a length and an area, what unit would you use to measure the sum?) and the same with its inverse subtraction. So even if you break up (a + b)^10 into a sum of other terms all the terms must still be the measure of a 10 dimensional “rectangliod”. This means they themselves are the product of 10 numbers multiplied.

You are also assuming the events to be independent. If you assume all events have equal likelyhood of occuring then you can find the probabilty of each outcome by dividing like you said. This is a way of working out all possible outcomes of a set of identical idepenent experiments?

When you generalise this to finding the outcomes if the experiments are not the same e.g roll a dice the flip a coin would be (a+b+c+d+e+f)*(x+y) the multiplication analogy holds (thousands etc.).

However if when you rolled the head matters then you loose a lot of the worth of this model (as you would be able to make the multiplication commute) and you are better of with the vector model. This is when you map each outcome, e.g. head, to a number on the cooridante axis. Then outcomes after say n experiments are all the points in R^n space. This is essentially what you are doing with the grid model.

The difference between combinations and multiplications are that the combinations of points in space (e.g (heads, tails) = (0,1) ) but multiplication is finding the area. If you were thinking about the probability of the events that would be the same as multiplication (at least with independent events) as it could be thought of a area in the same way.

Hey Kalid,
Great work :slight_smile:
one quesion: In how many ways can a batsman score 200 hitting sixes and fours only…

if he hits x fours and y sixes
4x + 6y=200
can we solve this equation using permutation and combination??? or do we have to go by the method of actually counting the ways of sixes and fours…


“Even a wrench can drive nails, once you understand the true nature of “being a hammer” (very Zen, eh?).” Corollary …even a hammer can remove nails once you understand the true nature of being a lever.

Hi, I think I can answer your question. It looks simpler if you look at it as this:
What’s the max. number of sixes that can be hit? 32. (since that makes it 32*6 = 192, from where we can arrive at a perfect 200 by adding 4s). To make it a perfect 200 each time, you go on replacing two 6s with three 4s, which can be done 16 times.

@Charles (comment #10):
I think your problem is that if you choose an element from one set, say A, you can’t choose only A from the other, while being forced to choose every other element. This means you begin with choosing a single element from either set and four on four of the other + two from either and three on three from the other, an so on…

It boils down to {2 (5C14C4 + 5C23C3 + 5C32C2 + 5C41C1)}. Or,
Simply, 2 (5 + 10 + 10 + 5) = 60 choices.

[…] is often seen as AND. Component A must be there AND Component B must be there. If either condition is false, the system […]

@mark: Whoa… very nice :).

@soni…the equation is

6C2+6C3+6C4+6C5+6C6…solving you will get 57…

I have a question I would to have answered if anyone of is able to. I need to know how many different combinations there are for two colours red and black five rows accross. eg: Red red red black black
Red Black red black black etc etc.
If anyone can help that would be much appreciated.

Thank you very much Kalid! :slight_smile:

how can we solve dice problems with help of equation?


Is there a chance for you to write an intuitive post on the Bayes’ Theorem?

Something along the lines of this article:

Your explanations are great!

Hi Carlos, I wrote about that one here:

Glad you’re enjoying the site!

What is the reasoning behind “and” becoming multiplication and “or” becoming addition?

Nevermind, I found an intuitive argument: if you imagine the total probability as a square, the probability of T AND T happening is the area of the region which is the intersection of the two segments representing T.

Great insight! I like it.

Opened up my mind a whole a lot thank you…