Re: How many Combinations ?

Been a long time I haven't revised my lessons. My Insight may be a little poor right now. But the question appears quite easy.

**(1)** If **order matters**, for ex. AB, BA counts as two combinations.

**10 numbers** (1,2,3,4,5,6,7,8,9,0) to be inserted into **2 slots**:

=$\displaystyle 10P2$

and

**26 alphabets** (a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z) to be inserted into the **5 slots**:

=$\displaystyle 26P5$

Finally, linking the two events together **(By Product Principle law)**:

**=$\displaystyle 10P2\times 26P5$**

**(2)** If **order don't matters**, then

**=$\displaystyle 10C2\times 26C5$**

Re: How many Combinations ?

Hi Zikcau25,

Thank you for responding, I appreciate the time that you have given to my question.

However,

I am sorry but I don't think that what you have written is correct.

10P2 gives the number of ways 2 items can be chosen from 10 items where order matters.

However, 2 distinct items must be chosen without replacement.

My question has replacement. That is, both of the numbers can be the same.

Therefore the number of ways a two digit number can be made is $\displaystyle 10\times10=10^2$ not $\displaystyle ^{10}P_2=10\times9$

Re: How many Combinations ?

HI;

My understanding is that with combinations you have a set of things and choose so many

things from that set ie you have 7 items lets say you choose 2, 7c2 = 21.

If your choosing all the items then your looking for the arrangments of those items

in which we need to look at factorials(!) ie you have 7 items 7! = 5040.

Hope this helps.

Re: How many Combinations ?

Quote:

Originally Posted by

**Melody2** If 2 numbers and 5 letters are chosen in any order, how many possible ordered combinations are there?

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It is definitely time for a vocabulary lesson.

Both words *combinations and permutations* imply **distinct objects**.

Combinations are __sets__ of objects. Permutations are __strings__ of distinct objects.

So I agree with relpy #2. Given the way you put the question the answer is $\displaystyle \binom{10}{2}\binom{26}{5}$.

Quote:

Originally Posted by

**Melody2** **My question has replacement**. That is, both of the numbers can be the same.

Therefore the number of ways a two digit number can be made is $\displaystyle 10\times10=10^2$ not [tex]^{10}P_2=10\times9[/tex

However, given the above reply, you are asking about multi-selections.

I don't care for the notation used on that webpage. But if you read it carefully the idea is to select N objects from K different types.

Given twenty-one flavors of ice cream and we select three scoops, we could have all three the same flavor or two of one and the third a different. **Note that no order is implied here**.

The number of ways to select N objects from K different types is $\displaystyle \binom{N+K-1}{N}.$

The answer to your question is $\displaystyle \binom{11}{2}\binom{30}{5}$.

Re: How many Combinations ?

Hi all

First I would like to offer an appology to Zikcau25. I did question my use of the word 'combination' but I didn't think of the term 'selection' and I thought I had seen combination used loosely like this somewhere else. I thought my initial answer would help overcome any misinterpretation but i realize now that this was an unreasonable assumption and that considering how poorly my question was written, your combination answer was reasonable. Please accept my appology.

My thanks also to Anthonye. I appreciate your answer.

Now Plato,

Thankyou for your vocabulary lesson. I don't think that I will forget the word 'selection' or how it is used for a very long time. lol.

Your lesson on selecting N objects from K different types was very welcome. I read the article on multi-selection and I am very please to have added that data to my memory bank.

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I know that the number of combinations that I can get if I choose 2 numbers from 10 is 10C2 and

I know that the number of combinations that I can get if I choose 5 letters from 26 is 26C5 and

I know that the number of combinations that I can get if I choose 2 numbers from 10 and 5 letters from 26 is 10C2 x 26C5

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I now know that the number of selections (where order doesn't count) that I can get if I select 2 numbers from 10 is (2+10-1)C2 = 11C2 and

I now know that the number of selections (where order doesn't count) that I can get if I select 5 letters from 26 is (5+26-1)C5 = 30C5 and

I now know that the number of selections (where order doesn't count) that I can get if I select 2 letters from 10 and 5 letters from 26 is 11C2 x 30C5

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I know that the number of permutations that I can get if I choose 2 numbers from 10 is 10P2 and

I know that the number of permutations that I can get if I choose 5 letters from 26 is 26P5 and

I know that the number of permutations that I can get if I choose 2 letters from 10 and 5 letters from 26 (Where the numbers must be presented first and the letters afterwards) is 10P2 * 26P5

However,

**I DON'T know** what the number of permutations that I can get if I select 2 letters from 10 and 5 letters from 26 and the letters and numbers can be muddled up in any order. I am thinking that it might be 10P2 * 26P5 * 7!/(2!5!)

(I think that this might be correct because for each possible permutation the order of the numbers in relation to the other number is already allowed for and the order of the letters in relation to the other letters is already allowed for. only the order of the position of the numbers in relation to the letters is not yet accounted for.

So, for each permutation of 5 letters and 2 numbers, there are 7!/(2!5!) ways of ordering them together.

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**NOW my question (as I intended it) is different from all of the above. the question is;**

**If 2 numbers and 5 letters are selected in any order, (with replacment) how many possible ordered selections are there?**

My question is like a permutation question EXCEPT each letter or number can be chosen more than once.

If the two numbers came first and the five letters followed, I believe it would be $\displaystyle 10^2\times26^5$

Now I can think in terms of just 2 numbers and 5 letters and I think there might be 7C2 that is 7! / (2!5!) ways of intermixing each ordered selection of numbers with each ordered selection of letters.

So I am thinking that the answer might be $\displaystyle 10^2\times26^5\times 7C2$

So, I am still exactly where I started.

I am definitely NOT sure that this is correct, but of all the possiblities that I have thought of, or that have been presented to me. I think it is the most likely one to be correct.

ANYONE up for more discussion or teaching. I'd really like to get an answer that I am completley comfortable with.

Thankyou.

PS I hope that i don't need any more vocabulary lessons on this question. (Worried)

Re: How many Combinations ?

Hi Melody.

You invite further discussion, so here's my take on the question.

When I first looked it, I arrived at your answer of $\displaystyle 7C2*10^{2}*26^{5}.$

This was my reasoning.

Denoting the numbers by N and the letters by L, (none of them fixed for the moment), I can write down the following possible strings,

N N L L L L L,

N L N L L L L,

N L L N L L L,

...

...

L L L L N L N,

L L L L L N N.

There are $\displaystyle 7C2 = 21 $of them in all.

Looking at that first string, **and assuming that the same number or letter can be used more than once**, the number of possible variations would be $\displaystyle 10*10*26*26*26*26*26=10^{2}*26^{5}.$

It's the same number of variations for each string, so that gets you the answer I gave earlier $\displaystyle 7C2*10^{2}*26^{5}$.

If a particular number or letter can be used once only, the number of variations for each string would be $\displaystyle 10*9*26*25*24*23*22= 10P2*26P5$ and therefore a total of $\displaystyle 7C2*10P2*26P5.$

Plato's (equally valid) interpretation of the question is different.

In essence he has two numbers at one end of the table and five letters at the other, the two groups not being brought together to form a string.

Re: How many Combinations ?

Quote:

Originally Posted by

**BobP** Plato's (equally valid) interpretation of the question is different.

In essence he has two numbers at one end of the table and five letters at the other, the two groups not being brought together to form a string.

Actually, I said the the answer I gave counts the number of possible selections that consist of two letters and five digits in which repetitions are allowed. I did say in that post that answer **does not imply **__order__.

@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ @

**Now if we **__change__ the question: How many strings that consist of two letters and five digits, repetitions are allowed, are possible?

@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

That is a very difficult question to answer. Let me give several examples to illustrate the difficulty.

Consider the selection $\displaystyle \{(N,L,4,4,1,1,5)\}$ there are $\displaystyle \frac{7!}{(2)^2}$ ways to arrange that selection into a string.

Consider the selection $\displaystyle \{(A,A,4,4,4,1,5)\}$ there are $\displaystyle \frac{7!}{(2)(3!)}$ ways to arrange that selection into a string.

Consider the selection $\displaystyle \{(N,L,1,2,3,4,5)\}$ there are $\displaystyle 7!$ ways to arrange that selection into a string.

I hope you can see what a nightmare it would be to count all the all the possible configurations.

Re: How many Combinations ?

Quote:

Originally Posted by

**Plato** @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@ @

**Now if we **__change__ the question: How many strings that consist of two letters and five digits, repetitions are allowed, are possible?

@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

Actually, I have changed my view of that question. Over my dinner break, it occurred to me that there is a simple solution.

There are $\displaystyle \binom{7}{2}$ ways to place the letters. There are $\displaystyle (26)^2$ ways to select the two letters.

So the answer is $\displaystyle \binom{7}{2}(26)^2(10)^5$

Re: How many Combinations ?

Hi Plato,

Quote from Plato "Actually, I said the the answer I gave counts the number of possible selections that consist of two letters and five digits in which repetitions are allowed. I did say in that post that answer __does not imply order__."

YES I know/knew that, I was not trying to cause any offense. I saw that in you post. I am sorry. I'm not out to offend anybody.

Thank you. I really appreciate all your answers. I am glad that i finally worded my question properly. If I had worded it properly in the first place I know there would not have been any problems.

Thank you.

Melody.