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**Hannibal lecter****Member**- Registered: 2016-02-11
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Hi, what is the relationship between (Combinations and Permutations) AND ( Probability) ????

can anyone help me with that please!!!

Wisdom is a tree which grows in the heart and fruits on the tongue

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**bob bundy****Administrator**- Registered: 2010-06-20
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hi Hannibal lecter

Some probability problems will require one or other of these.

For example: How many ways can you make a five letter 'word' (it doesn't have to be in the dictionary) from the letters of MATHEMATICS. What is the probability that the word has two Ms in it.

There are 11 letters and we are choosing 5, so that's the number of permutations of 5 from 11.

Now consider the words made. Some will have no Ms, some will have 1 M and some will have 2 Ms. So consider choosing 3 non M words first, then add two Ms but bearing in mind the Ms may be either way round (M1 before M2 or M2 before M1). That's a combination question because order of Ms doesn't matter.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Hannibal lecter****Member**- Registered: 2016-02-11
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can you give me another easy example please, I didn't understand that one, maybe an example of marbles or dice or coins, ..

please just give me an easy example,

and what did you mean when you said 5 of 11 ? is it mean p(n=5,k=11) or vice versa?

and did you use the rule of Permutations with Repetition or without ?

as well as ..

the rule of probability is = Number of ways it can happen (event) /Total number of outcomes( sample space)

is the (Combinations and Permutations) the same as =Total number of outcomes( sample space)? or what..

please help me with that one

*Last edited by Hannibal lecter (2018-03-09 00:14:01)*

Wisdom is a tree which grows in the heart and fruits on the tongue

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**bob bundy****Administrator**- Registered: 2010-06-20
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hi Hannibal lecter

I'll have a go with dice. Let's say you have three dice, numbered 1-6 as usual. And let's say the faces have the actual digits rather than dots. If you throw the first, then the second, placing it to the right of the first and finally the third placing it to the right of the second, you will be looking at a number with hundreds, tens and units.

What's the probability the number is 241 ?

The first number has to be a 2 (P= 1/6) the second a 4 (P=1/6) and the third a 1 (P=1/6) so the probability of getting this number is P = 1/6 x 1/6 x 1/6 = 1/216

Now what about if we just throw all three at once, and we're allowed to re-arrange them to try and make that number. That's easier to get because any of the three can be the 2, the 4 and the 1. So 1, 2, 4 would be ok.

There are still 216 possible outcomes. But how many will give us those numbers in any order:

We could throw 124, 142, 214, 241, 412, or 421 and re-arrange to get our target number. 6 possibilities is the number of **combinations** of three digits ie. 3x2x1.

In my example it was easy to write out all six possibilities but if we had lots more digits to consider (supposing it were ten dice) so it's useful to know there's a formula to get us there quickly.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Hannibal lecter****Member**- Registered: 2016-02-11
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2 (P= 1/6) the second a 4 (P=1/6) and the third a 1 (P=1/6), is the number gender is the 6 possibilities or the 6 digit of the die?

we didn't use the combinations , you just solve the question before calculating the combinations,

and now if we allowed to re-arrange them and we know that there is six possibilities what after?

Wisdom is a tree which grows in the heart and fruits on the tongue

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**bob bundy****Administrator**- Registered: 2010-06-20
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is the number gender is the 6 possibilities or the 6 digit of the die?

These are the same thing. You consider how many equally likely events there are. The dice has six faces. There are six numbers. Either one gives 1/6.

If you are allowed to re-arrange the order of the throws there are six ways of getting 241. Altogether there are 216 possibilities. So the probability of throwing that combination is 6/216

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Hannibal lecter****Member**- Registered: 2016-02-11
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How did you calculate this combinations by its rule?

or you just know everything about the dice

please tell me how did you calculate this using the rule of combinations (n!/r!(n − r)!)

Wisdom is a tree which grows in the heart and fruits on the tongue

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**bob bundy****Administrator**- Registered: 2010-06-20
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This site has many good examples:

https://courses.lumenlearning.com/finit … binations/

It also looks at the birthday problem, taking it slowly with easier cases first.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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