Combinations

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Fudge111211111

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Players select 5 numbers between 1 and 50
And 2 additional numbers between 1 and 11. Players can select same number more than once. How many different combinations of selection these 7 numbers? Give ans in standard form
——the answer is 3.78*10^10 but how do I get that answer??
 
Hi, on this forum we will not tell you how to get to the correct answer. However we will help you arrive at the correct answer. However you must tell us where you are stuck. By showing us something then we know which method you want to use and then we can guide you using that method to the correct answer.
 
In choosing 5 numbers between 1 and 50 can you choose the same number? When you choose two numbers between 1 and 11 can you choose the same number? Or maybe when you choose to numbers between 1 and 11 it can be the same number as from 1 to 50?

Before you can do this problem you must know what it means when it says that you can choose the same number.
Based on your answer you can repeat a number when you choose 5 numbers from 1 to 50 AND when choosing 2 numbers from 1 to 11. So you can, if you like, choose a number 7 (from 1 to 11) times.
 
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Fudge111211111 said:
Players select 5 numbers between 1 and 50
And 2 additional numbers between 1 and 11. Players can select same number more than once. How many different combinations of selection these 7 numbers? Give ans in standard form
——the answer is 3.78*10^10 but how do I get that answer??
Click to expand...
Comment: I think that you mean From 1 to 50? Between means neither 1 nor 50 can be included.
There are \(\dfrac{(50-1+5)!}{(49!)(5!)}=3162510\) ways to select five integers from one to fifty.
There are \(\dfrac{(11-1+2)!}{(10!)(2!)}=66\) ways to select ftwo integers from one to eleven.
The product of those two is the answer.
 
66 ways to choose two integers from one to eleven? I wonder about that.
 
Jomo said:
66 ways to choose two integers from one to eleven? I wonder about that.
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What do you wonder.
There are \(\dbinom{11}{2}=55\) ways to select two distinct integers from one to eleven.
There are \(\dfrac{(11-1+2)!}{(10!)(2!)}=66\) ways to select two integers from one to eleven.
 
Fudge111211111 said:
Players select 5 numbers between 1 and 50
And 2 additional numbers between 1 and 11. Players can select same number more than once. How many different combinations of selection these 7 numbers? Give ans in standard form
——the answer is 3.78*10^10 but how do I get that answer??
Click to expand...

Clearly, judging by the provided answer, this is meant to be a very simple permutation problem; the word "combinations" is not to be taken literally, because order does matter. That's unfortunate; I wonder if the OP paraphrased the problem? (The grammar suggests that.)

EDIT: I don't literally mean permutation, but just that order matters.

If it were actually about combinations with repetition, it would require a considerably higher level of thinking, which is what pka is doing.

Perhaps the best thing to do is to restate the problem in terms of something like filling in a form, where the first 5 blanks will hold a number from 1 to 50, and the last two hold a number from 1 to 11. How many ways can it be filled in?
 
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Otis said:
Because there are eleven ways to pick the first integer and eleven ways to pick the second integer. That's 112 ways total.
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That is then wrong counting model. These are multi-sets. How can we pick two integers from one to eleven, allowing reputations.
Now you are correct that there are \(121\) ordered pairs in \(\{1,2,3,4,5,6,7,8,9,10,11\}\times\{1,2,3,4,5,6,7,8,9,10,11\}\)
But the pairs \((2,6)~\&~(6,2)\) are the same selection of \(2~\&~6\).
This the correct model counts the number of ways to place two identical objects(choices) into eleven distinct cells(positive integers).
The number of ways to place \(N\) identical objects into \(D\) distinct cells is \(\dbinom{N+D-1}{N}\)
 
OK, I do not understand how I did not see how pka was getting his result. It was obvious but I was just looking at 11^2 and nothing else (after all 11^2 is correct)
 
nope, I meant between 1 to 50
The question is directly from a edexcel workbook, I tried many ways but I can’t get the answer of 3.78*10^10
 
Jomo said:
In choosing 5 numbers between 1 and 50 can you choose the same number? When you choose two numbers between 1 and 11 can you choose the same number? Or maybe when you choose to numbers between 1 and 11 it can be the same number as from 1 to 50?

Before you can do this problem you must know what it means when it says that you can choose the same number.
Based on your answer you can repeat a number when you choose 5 numbers from 1 to 50 AND when choosing 2 numbers from 1 to 11. So you can, if you like, choose a number 7 (from 1 to 11) times.
Click to expand...
I really don’t understand- in the textbook the topic is combinations, all it says on that page is to multiply to get number of combinations. At the bottom of that page is that question. I tried multiplying all key numbers but I can’t get that answer. So I don’t know
 
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Jomo said:
OK, I do not understand how I did not see how pka was getting his result. It was obvious but I was just looking at 11^2 and nothing else (after all 11^2 is correct)
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Do you know how to work get that final answer written in my question ? I don’t get it ):
 
Fudge111211111 said:
nope, I meant between 1 to 50
The question is directly from a edexcel workbook, I tried many ways but I can’t get the answer of 3.78*10^10
Click to expand...
Fudge111211111 said:
Do you know how to work get that final answer written in my question ? I don’t get it ):
Click to expand...

All they did is to multiply 50*50*50*50*50*11*11 = 50^5 * 11^2 = 37,812,500,000.

This makes sense if you interpret the problem as I did in post #11, last paragraph, and do what I said in post #2. Each choice you make is a multiplication; each of the first 5 choices is among 50 possibilities, and the last two are each among 11.

The difficulty is that the problem is very poorly worded. If you quoted it exactly, then the workbook needs a lot of improvement. But also, the wording might make more sense to us if we saw the context (for example, if they hadn't just defined what "combinations" means.
 
Dr.Peterson said:
All they did is to multiply 50*50*50*50*50*11*11 = 50^5 * 11^2 = 37,812,500,000.
Click to expand...
If that is correct, the question is indeed poorly worded. If player were handed a blank card with seven boxes as below,
\(\boxed{\color{white}{|\_\_\_\_}}\;\boxed{\color{white}{|\_\_\_\_}}\;\boxed{\color{white}{|\_\_\_\_}}\;\boxed{\color{white}{|\_\_\_\_}}\;\boxed{\color{white}{|\_\_\_\_}} \;\;--\;\boxed{\color{white}{|\_\_\_\_}}\;\boxed{\color{white}{|\_\_\_\_}}\)
to fill in; the first five with any integer from one to fifty and the last two five with any integer from one to eleven then 50^5 * 11^2 is indeed the answer.
If this is from a published text, it was not reviewed carefully. The OP clearly says that a player selects. Selections are not ordered, unless the problem clearly calls for order as with an election for club offices. Winning lottery numbers are ordered numerically without regard to the order in which they were selected. Selections and combinations are content driven as apposed to permutations.
 
Agreed!

According to posts #14 and #15, this is indeed from published material, presumably copied word for word. My only reason for supposing it makes sense to its authors is that it is probably very introductory material on combinatorics, before much has been taught, apparently even before the technical meaning of "combination" has been mentioned.

And in fact, having said that, I searched for the exact wording and found it: https://www.pearsonschoolsandfecoll...igher/edexcel-gcse-maths-higher-rg-sample.pdf

See page 18 of the PDF (page 13 as printed). Though I don't see the answer given, and the problem wasn't copied exactly (the word "combinations" is not used in the problem, though it is used with the same wrong meaning at the top of the page), it's the same problem with the same abuses of terminology. It's at least close to mathematical malpractice.
 
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