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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
10 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
10 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
10 years ago

Wats the ans,is it 256?

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