How to solve such problem?

Question

How many solution exist for x+y+z=10 with 1<=x<4 1<=y<=7 and z>1

Answer 1

Ans will be 18

for x=1, y=1, z=8

             y=2 , z=7

.......................

            y=7 , z=2          => Total 7 numbers

 

for x=2, y=1, z=7

             y=2 , z=6

.......................

            y=6 , z=2        => Total 6 numbers

for x=3 , y=1 , z=6

              y=2,   z=5

 ............................

            y=5 ,  z=2           => Total 5 numbers

So, no of solutions 7+6+5=18

Comments

Any generalized solution or method if range of values for x are more?

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@srestha i am getting 22 using generating function .. can you verify once

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means in question like "no of solution exists" u just put it in generating function?

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here 2 variable has upper and lower constrain so its easy to calculate by generating function ... rather than doing by general method  i think

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@sid1221 can u show ur calculation?

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22 is ans

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22 is ans

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@sid1221
I think you did some mistake while calculating. Because by using generating function I got.
$\binom{8}{2}$ - $\binom{5}{2}$ = 28 - 10 = 18.

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may be possible , i have to check  once

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@Sid 121.There is a mistake in your initial conditions.

The constraint on x is :- 1<=x<4

Less than 4 means it can take maximum 3 value.But your solution takes the value as 4.

then you will get (1-x^3) instead of (1-x^4).

So required answer will be C(8,6)-C(5,3) =18.

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Ohh my bad I copied wrong question ...thanks for pointing out

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I cannot understand

@sid1221

why u taken $\frac{1-x^{4}}{(1-x)^{3}}$

instead of $\frac{\left ( 1-x^{4} \right )\left ( 1-x^{7} \right )}{\left ( 1-x \right )^{3}}$

@Hemant can u give ur solution plz

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sreshta we have to find x^6  coeifficrnt so (1-x⁴)(1-x⁷) = 1- x⁷(which  is more power consider as 0) -x⁴+ x¹¹(this one also not considered )

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@srestha