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phaelan.vaillancourt
Students
 
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Joined: Sat Apr 20, 2013 11:06 pm
 

Word Problem Combinatorics

by phaelan.vaillancourt Tue Jul 09, 2013 10:35 pm

Word Problem Guidebook, Chap. 10 Hard Question Set, last problem (believe its #20). Problem asks to calculate total amount of 3 digit numbers with the following conditions: 1) all 3 digits cannot be the same integer, 2) 0 cannot be a digit.

ManhattanGRE solution is 9x9x9 minus 9 (all digits from 111 to 999 that have all 3 digits the same integer) = 720. This makes sense to me however I don’t understand why 9x9x8 is not a solution. In other words, 9 (digits 1-9) x 9 (digits 1-9) x 8 (less one digit so that 3rd digit does match 2 other digits). For example, a menu has 2 appetizers, 2 mains, 2 deserts. The total possible combinations is 2x2x2 = 8. And if you just are allowed one dessert then it is 2x2x1 = 5. I’m not sure why the same logic doesn’t apply to digits question above. Thank you.
tommywallach
Manhattan Prep Staff
 
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Re: Word Problem Combinatorics

by tommywallach Thu Jul 11, 2013 9:21 pm

Hey Phaelan,

I'd prefer the slots method, too, but you made one mistake! The third slot can be 9 things IF the second slot didn't match the first slot. So you can't just put an 8 in the third slot, because your choices for the third slot DEPEND on what happens in the first two slots. So you'll need to separate out these two possibilities and add them together.

Number of ways to have the first two numbers the same and third different.

First two slots same = 9 ways (11, 22, 33, etc.)
Third slot = 8 (no zero, and can't be the same as first two)
9 * 8 = 72

Number of ways to have first two numbers different (meaning the third number is free to be whatever):

First slot: 9 numbers (everything but zero)
Second slot: 8 numbers (everything but zero, and whatever's in the first slot)
Third slot: 9 numbers (everything but zero)
9 * 8 * 9 = 648

Finally, we add it all up:

648 + 72 = 720

Hope that helps!

-t