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Combination Type Formula Shortcut

Expert replies
by EMAN » Thu Sep 10, 2009 8:07 pm
There was a problem asking how many flights between a pair of 50 cities were possible. Here is a shortcut I thought would be helpful, although I'm sure there's a single equation somewhere. To solve it, you would have to find out 50+49+48+47+46...etc.

50 x 10 = 500 (50, 51, 52, 53, 54, 55, 56, 57, 58 , 59) - just ignore the one's digit here

40 x 10 = 400
30 x 10 = 300
20 x 10 = 200
10 x 10 = 100

Now, the one's digit in each of these sets is always 45. 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45

45 x 5 sets = 225

1-9, 10-19, 20-29, 30-39, 40-49 (5 sets)


Grand Total = 1,725
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Source: — Problem Solving |

by Nermal » Fri Sep 11, 2009 6:36 am
I think this approach is very intuitive, thank you!

I think there is something wrong though:
50*10 would assume that you have numbers greater than 50.

You have 5 sets, as you said:
0-9: 0+1+2+...+9=45
10-19: 10*10+45
20-29: 20*10+45
30-39: 30*10+45
40-49: 40*10+45

Hence there is a total of 1225 possible flights.
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Re: Combination Type Formula Shortcut

by Ian Stewart » Fri Sep 11, 2009 7:17 am
EMAN wrote:There was a problem asking how many flights between a pair of 50 cities were possible.
Notice that you can also solve this problem using counting principles. It's not clear whether the order should matter (without seeing the wording of the original question); that is, do we consider a flight from LA to Chicago to be different from a flight from Chicago to LA? If so, then when choosing a flight we have 50 choices for the first city and 49 for the second, so 50*49 = 2450 flights in total. If order is not important, then we need to divide by 2, so we have (50*49)/2 = 1225 flights (this is sometimes written as 50C2, the number of ways of choosing 2 things from 50 if order does not matter).

Note also that if order is not important, this is, mathematically speaking, the same problem as "50 people at a party all shake hands. How many handshakes take place?", or "In a tennis tournament, 50 players each play exactly one match against each other opponent. How many matches are played?", or "50 distinct dots are drawn on a page. How many lines can be drawn which connect pairs of dots?" See Q121 in the PS section of OG12 for a similar example.

There are a lot of counting problems that look different on the surface, but all boil down to the same basic idea.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com
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Follow Up

by EMAN » Fri Sep 11, 2009 8:41 am
Nermal - That's correct. I was mistaken with the 50's section. Thank you for pointing that out.

Ian - Thanks for the excellent commentary. That is very helpful.
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nCr

by EMAN » Fri Sep 11, 2009 8:51 am
By the way Ian, were you refering to the nCr combination formula above if order does not matter? I assume it goes something like:

50! / 2! (50 - 48)! which comes to 50 x 49 = answer / 2!
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