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Test Series ACE

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1 1 vote
How to solve?
________________

Number of non-negative integer solutions such that

x+y+z=17

where x>1,y>2,z>3 is --------

2 Answers

0 0 votes
since x>1,y>2 and z>3

so new equation will be

x+1+y+3+z+4=17

=>x+y+z=8

now no. of solution of these equations are (n+r-1)C(r-1)

here n=8 r=3(no. of unknowns)

so 10C2=45 Ans
0 0 votes

Another way to think is by the constraint method..!!

See..
Let's find the max value of z, that could fit the equation..!
For that x and y must be least...{2, 3 respectively)
So, max(z) = 12..

So, 4 <=z <=12..
Similarly, 2 <= x <=10..
And, 3 <= y <= 11
Now, for z=4, there are 9 solutions possible (for each value from y=3 to 11, there will always be 1 value of x.)
For z=5, there are 8 solutions possible (from y=4 to y=11, there will be a value of x, but for y=3, there's no value for x.)

This goes on, till z=12, there will be only 1 solution...@ Simple pattern will be seen..!

So, the total solutions will be..
9+8+7+6+5+4+3+2+1 = 45..!!

Method if not ready to go with the Formula based logic..!!

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