Consider x=1.

y+z = 8 (y can take values from 1 to 7)

Considering x =2

y+z = 7 (y can take values from 1 to 6)

You can keep going till x=7, where y=z=1.

So, all possible cases will be 1+2+3+... +7 = 7*8/2 = 28

knight247 wrote:If x, y, and z are positive integers, and x + y + z = 9. How many combinations of x, y, and z are possible?

OA 28

Hey Ashwin,
I'm really not able to understand ur solution Bro. Can u plz elaborate a little more. Thanks

@knight247,
given equation x+y+z=9 --(a)

First let us consider that x=1,
To satisfy the condition for the eqn a, y and z can assume the following values
(1,7),(2,6),(3,5),(4,4),(5,3),(6,2),(7,1) [read as (y,z)], when counted will be equal to 7 combinations (as x,y,z are only +ve integers)

similarly, if x=2, then the values of (y,z) would be
(1,6),(2,5),(3,4),(4,3),(5,2),(6,1) when counted will be equal to 6 combinations

If you observe the series, depending on the x value from x=1 to 7, then number of combinations would be 7+6+5+4+3+2+1 = 28

Hope u understand the solution now.

Regards
Chetan
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I think Chetans explanation is pretty detailed. Sorry I wasnt elaborate in my reply.

Think of it this way,

When x =1.

y varies from 1 - 7 (z can take only 1 value depending on which value y takes) So 7 cases.

Now when x =2

y varies from 1-6 (Again z can take only 1 value in each case depending on what value y takes) So 6 cases.

Similarly, for each incremental value of x, the number of values y can take reduces by 1.

Adding all the possibilities will give you 7+6+5+4+3+2+1 = 28.

In addition, if the sum mentioned x,y,z could take 0 also, then using the same logic, you would have 9+8+7+....+1 = 45 cases.

Similarly if the same sum was given to be x+y+z=10

You could directly do, 8*9/2 = 36 and so on.

Hope you understood.

Cheers.

Hi

I will give my level best in explaining.

Assume 9 boards with number 1 written on it separated using underscores...

1_1_1_1_1_1_1_1_1

if i select any two underscores randomly then those two spaces will partition above boards into three groups(x,y,z), with each group when summed up making 9(x+y+z).

And selection of these two spaces out of eight available is 8c2 = 28 ways

and you can just generalize this like...
if x+y+z = n, and x,y,z are +ve integers then formula is (n-1)C2

HTH!
user123321