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Saumyojit

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X*y*z=30
Find the no of positive integral solutions.
How will I use multinomial theorem here?


If, X+Y+Z=10 was given then I know how to use multinomial theorem to get solutions.




Prime factorization of 30=2^1* 3^1* 5^1

Now no of divisors is 8 .
Using multinomial theorem Approach I create a random variable c.
to the power of 'c' is will be every divisor of 30 possible.
Now,
(c^1+ c^2+c^3+c^5.....)
Total 8 terms as they're 8 divisors of 30
Now this set will appear thrice which are Alike as three variables are given x.y.z=30
Between each set there is a multiplication
Set1* Set2 * Set3
(Set)^3


The Series of each set are in GP ? Not sure
Sum of GP =[first term *(1-r^n)] / (1-r)
What is the common ratio?
 
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Well 2 have 3 options x y or z same for other numbers so answer is 3*3*3=27 hope it helps stemtrainers
 
I'm not at all sure I follow your thinking.

Can you show us where you got the idea of using the multinomial theorem, and perhaps an example from there of how it might be used in other problems of this sort? (Perhaps show your x+y+z=10 example.)

The way I see it, you want to count all ways to assign each of the factors 2, 3, and 5 to one of the three variables x, y, and z. For example, if 2 and 3 both go to x and 5 goes to z, you get (2*3)(1)(5), so x=6, y=1, z=5.
 
Dr.Peterson said:
I'm not at all sure I follow your thinking.

Can you show us where you got the idea of using the multinomial theorem, and perhaps an example from there of how it might be used in other problems of this sort? (Perhaps show your x+y+z=10 example.)

The way I see it, you want to count all ways to assign each of the factors 2, 3, and 5 to one of the three variables x, y, and z. For example, if 2 and 3 both go to x and 5 goes to z, you get (2*3)(1)(5), so x=6, y=1, z=5.
Click to expand...


watch the video from 21minutes
 
It would help if he spoke English. If he is, it's incomprehensible.

But why do you want to solve your current problem using this method? (I am not saying you can't, just that you haven't explained why you want to.) Did someone tell you to, or have you seen an example more like yours?

If you do use this method, you need to figure out how the answer to your problem can be thought of as a coefficient in a power or product of polynomials. Do you have a reason for the suggestion you made in the OP?
 
If you are studying for iit let me tell you this
1.You are wasting your precious time by coming to these forums; hire a professional teacher for doubts and clarity of concepts.
2. If not studying for iit then you can spend time here no problem.
I did my btech from iitbombay.
So believe me.
 
Saumyojit said:
X*y*z=30
Find the no of positive integral solutions.
How will I use multinomial theorem here?


If, X+Y+Z=10 was given then I know how to use multinomial theorem to get solutions.




Prime factorization of 30=2^1* 3^1* 5^1

Now no of divisors is 8 .
Using multinomial theorem Approach I create a random variable c.
to the power of 'c' is will be every divisor of 30 possible.
Now,
(c^1+ c^2+c^3+c^5.....)
Total 8 terms as they're 8 divisors of 30
Now this set will appear thrice which are Alike as three variables are given x.y.z=30
Between each set there is a multiplication
Set1* Set2 * Set3
(Set)^3


The Series of each set are in GP ? Not sure
Sum of GP =[first term *(1-r^n)] / (1-r)
What is the common ratio?
Click to expand...
Well there is logic problem here when you multiply the sets power add and in question the terms multiply so it doesn't make sense to do that.
 
Stemtrainers said:
Well 2 have 3 options x y or z same for other numbers so answer is 3*3*3=27 hope it helps stemtrainers
Click to expand...
you are saying that if you have fixed in your mind 2,3,5 triplets.
now you can arrange 2 in 3 ways .

what if the no is bigger (300)
then i cannot break down all the triplets.
 
Dr.Peterson said:
It would help if he spoke English. If he is, it's incomprehensible.

But why do you want to solve your current problem using this method? (I am not saying you can't, just that you haven't explained why you want to.) Did someone tell you to, or have you seen an example more like yours?

If you do use this method, you need to figure out how the answer to your problem can be thought of as a coefficient in a power or product of polynomials. Do you have a reason for the suggestion you made in the OP?
Click to expand...
He is speaking in HEnglish - Hindi+English !!
 
Saumyojit said:
what if the no is bigger (300)
Click to expand...
Find the prime factors...
300 = 2² * 3 * 5²

Then, using stars and bars...
How many ways to distribute the "2" factors within the three buckets x,y, and z: **|| [imath]\frac{4!}{2!2!} = 6[/imath]
How many ways to distribute the "5" factors: **|| [imath]\frac{4!}{2!2!} = 6[/imath]
How many ways to distribute the "3" factor: *|| [imath]\frac{3!}{2!} = 3[/imath]

Multiply these results together, 6*6*3 = 108 ways

EDIT: Please answer the following question...
Dr.Peterson said:
But why do you want to solve your current problem using this method? (I am not saying you can't, just that you haven't explained why you want to.) Did someone tell you to, or have you seen an example more like yours?
Click to expand...
...because perhaps you have been shown a particular method to solve this type of problem. And we would like to show you how to use THAT method rather than showing you a (potentially) different method. But, I suspect (from the video) that you're thinking of a method that can be used for addition. This same method can't be used for multiplication (as far as I'm aware).
 
Last edited:
Cubist said:
Find the prime factors...
300 = 2² * 3 * 5²

Then, using stars and bars...
How many ways to distribute the "2" factors within the three buckets x,y, and z: **|| [imath]\frac{4!}{2!2!} = 6[/imath]
How many ways to distribute the "5" factors: **|| [imath]\frac{4!}{2!2!} = 6[/imath]
How many ways to distribute the "3" factor: *|| [imath]\frac{3!}{2!} = 3[/imath]

Multiply these results together, 6*6*3 = 108 ways

EDIT: Please answer the following question...

...because perhaps you have been shown a particular method to solve this type of problem. And we would like to show you how to use THAT method rather than showing you a (potentially) different method. But, I suspect (from the video) that you're thinking of a method that can be used for addition. This same method can't be used for multiplication (as far as I'm aware).
Click to expand...
a great method . thanks
 
Saumyojit said:
just see the solution which is given by cubist @Stemtrainers .

@Stemtrainers sorry but you are misleading and you knew the answer 27 so you created your illogical approach to mislead .
Click to expand...
Don't be too hard on @Stemtrainers because their method works for the original question. The method I gave in post#11 is more general (it works in more cases). However, IF each factor appears only once then a shortcut exists:-

#ways = (the number of variables) ^ (the number of factors)

eg A: 30 =2*3*5 = x*y*z

Using the general method from post#11...
How many ways to place the "2" factor within the three buckets x,y, and z: *|| [imath]\frac{3!}{2!} = 3[/imath]
How many ways to place the "5" factor: *|| [imath]\frac{3!}{2!} = 3[/imath]
How many ways to place the "3" factor: *|| [imath]\frac{3!}{2!} = 3[/imath]

Multiply these results together, 3*3*3 = 27 ways

Using the shortcut, (the number of variables) ^ (the number of factors) = 3^3 = 27 ways

eg B: 30 = 2*3*5 = w*x*y*z so there are 4^3 = 64 ways

eg C: 50 = 2*5² = x*y this time we can't use the shortcut because one of the factors (5) appears more than once, it is squared. Therefore, using the general method, the answer is 6 ways.
 
This may be more that you want, if so just ignore.
The number N=452540000 in factored form is [imath]N=2^5 * 5^4 * 11^3 * 17[/imath].
Looking at the exponents only we see that [imath](5+1)(4+1)(3+1)(1+1)=(6)(5)(4)(2)=240[/imath].
This tells us the given number has two hundred and forty divisors. Why did that work and why add one?
Now the number [imath]5^k,~0\le k\le 4[/imath] is a divisor of N. That is five possible exponents.
So to find the number of divisors: factor into prime factors, add one to each power and multiply.
 
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