Combinations : Hard

Question

x,y,z are positive integers such that x+y+z = 100
Find the total combinations such that 25 15 & z > 20

shrouded1 · Answer

The total possible values for x are 25 (between 26 and 50, both inclusive)

We know y>15, and we also know that z is constrainted to be atleast 21. So as we increase y, the problem will only start occuring once x is so big that z has to be less than 21.

y = 16
x = 50 ... z = 34 (all lower x values are fine)
So 25 possibilities

y = 17 ... to ... 29 will also be fine by same logic
So 25 x 13 further possibilities

For y = 30 .... x = 26 to 49 works ... 24 possibilities
y = 31 .... x = 26 to 48 works ... 23 possibilities
.
.
.
.
y = 53 ... x = 26 only works ... 1 possibility
y > 53 ... 0 combinations work

So total possibilities there are = 1+2+...+24 = (24)(25)/2 = 12*25 = 300

So answer = 25 + 25x13 + 300 = 350 + 300 = 650

bagrettin · Answer

If x=26, then y+z=74. However y should be not less than 16, z should be not less than 21.
Then we could have y=16, z=58; y=17, z=57....y=53, z=21 - the total of 38 options.
If x=27, then y+z=73. Then we could have y=16, z=57....y=54, z=21 - the total of 37 options...
If x=28, then y+z=72. Then we could have y=16, z=56...y=53, z=21 - the total of 36 options...
......
If x=50, then y+z=50. Then we could have y=16, z=34...y=29, z=21 - the toal of 14 options.

Finally, overall number of combinations is 38+...+15+14=12*(38+15)+14=12*53+14=636+14=650

gmat1220 · Answer

x(max)=50 x(min) = 26 x = {26......50} total values = 25
y(max)=53 y(min) = 16 y = {16.......53} total values = 38
z(max)=58 z(min) = 21 z = {21......58} total values = 38

Total combinations = 25 * 38 * 38

Not sure !

bagrettin · Answer

shrouded1, there are 13 numbers between 17 and 29, therefore you should calculate 25*13.
Then our answers will be the same)

shrouded1 · Answer

agreed ... 650 it is !

gmat1220 · Answer

Hey bagrettin Just thinking why its not 25*25. i.e. for every value of x, I have a value of y and z such that y + z = 100-x ? This is a very lengthy calculation for every x. Your solution is kick ass ;-)

lawschoolsearcher · Answer

Wow, this problem is long and ugly for me. Do they really ask this on the GMAT????

My Solution:
x ranges from 26 to 50

So, if x=26, y will definitely range from 16 to 53
53-16 + 1 = 38 combinations of y and z

So, if x=27, y will def. range from 16 to 52
52-16 + 1 = 37 combinations of y and z

So, if x=50, y will def. range from 16 to 29
29-16+1 = 14 combinations

There is a pattern of 38 + 27 + .. + 14 combinations...

Let's use the Average = Sum of terms x n terms:
Average = 14 +38 /2 = 26
Terms = 38 -14 +1 = 25

Sum = 26x25 = 650!

bagrettin · Answer

Hello, gmat1220
It is not so because for each value of x you have different number of combinations of y and z which sum gives 100. This is due to restrictions on y and z. As you could see, the more is x the less options there are for y and z.
I am happy that you like my solution!

gmat1220 · Answer

bagrettin Just one word .... awesome +1, your solution amazes me ! I want this kind of thinking on gday ;-)

bagrettin · Answer

gmat1220, thank you)
I'm confident you will be successful on GDay!!!
And I wish I had the same kind of thinking in my verbal part)

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Combinations : Hard

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