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The no. of +ve integral solutions of x 1 x 2 x 3 x 4 =210 is=?

The no. of +ve integral solutions of x1x2x3x4=210 is=?

Grade:12

3 Answers

Ashwin Muralidharan IIT Madras
290 Points
9 years ago

Hi Basit,

 

Fist factorise 210 = 2*3*5*7.

 

Now, this has no "1".

 

When you want to have one "1". Then from the above 4 factors, pick two and club it. Which can be done in 4C2 = 6 ways.

 

When you need to have two 1s, pick three and club = 4C3 = 4 ways.

 

When you need to have three 1s, club all = 1way (ie 1*1*1*210)

 

And finally without 1s, it is the same as above 2*3*5*7.

 

Now, if you need to find the number of solutions, for the first case, the 4 distinct factors can be futher arranged in 4! ways.

So 4C2*4!

 

Similarly, second case, where two ones are there 4C3*(4!/2!)

 

And third Case 1*(4!/3!)

 

And fourth case again 4! (all are distinct)

 

Add all these cases, and you will get the number of solutions as: 24*6 + 4*12 + 4 + 24 = 220 solutions.

 

Hope that helps.

 

Best Regards,

Ashwin (IIT Madras).

 

Ashwin Muralidharan IIT Madras
290 Points
9 years ago

 

Hi Basit,

My Previous solution had an error in one particular case (Pls ignore)....

 

First factorise 210 = 2*3*5*7.

 

Now, this has no "1".

 

When you want to have one "1". Then from the above 4 factors, pick two and club it. Which can be done in 4C2 = 6 ways. (an example would be 1*6*5*7)

 

When you need to have two 1s, pick three factors and club = 4C3 = 4 ways, or club any two factors and the remaining two factors into another = 4C2 ways. An Eg (ie 1*1*30*7 or 1*1*6*35)

 

When you need to have three 1s, club all = 1way (ie 1*1*1*210)

 

And finally without 1s, it is the same as the original factors 2*3*5*7.

 

Now, if you need to find the number of solutions, for the first case, the 4 distinct factors can be futher arranged in 4! ways.

 

So 4C2*4!

 

Similarly, second case, where two ones are there (4C3+4C2)*(4!/2!)

 

And third Case 1*(4!/3!)

 

And fourth case again 4! (all are distinct)

 

Add all these cases, and you will get the number of solutions as: 24*6 + 10*12 + 4 + 24 = 292 solutions (if order of x1,x2,x3,x4 is important....)

If order is not imoprtant: (It would be 6+10+1+1 = 18 solutions)

 

Hope that helps.

 

Best Regards,

Ashwin (IIT Madras).

Sathya
35 Points
9 years ago

Wats the ans,is it 256?

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