Combinatronics, MGMAT

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rb90: Senior | Next Rank: 100 Posts; Posts: 77; Joined: Mon Jul 12, 2010 12:14 am; Location: India; GMAT Score:710

by rb90 » Thu Oct 14, 2010 2:06 am

The OA is [spoiler]20. I am getting 10 thriugh 2C1*5C4.[/spoiler]
Please explain where i am going wrong.
Thanks in advance guys.

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Source: Beat The GMAT — Problem Solving | Thu Oct 14, 2010 2:06 am

Geva@EconomistGMAT: GMAT Instructor; Posts: 905; Joined: Sun Sep 12, 2010 1:38 amFollowed by: 123 members; Thanked: 378 times; GMAT Score:760

by Geva@EconomistGMAT » Thu Oct 14, 2010 2:42 am

Not sure how OA is 20. I get 126.

What you have here are different arrangements of 5 steps of Down (D, D, D, D, D) and 4 steps of Left (L,L,L,L,).
so a single rout could be D, D, D, D, C, L, L, L, L, or L,L,L,L, D, D, D, D, D (which are the two routes along the sides of the rectangle) or any combination of these.

How many ways are there to arrange these 9 letters? If we had 9 different letters, we'd go 9!. However, the Ds and the Ls are the same internally, so the internal order of rearranging the Ds around in their positions for example, should be discounted. You basically need to divide by the number of members in each group factorial. The end result is 9! / 5!4! = 9*8*7*6 / 4*3*2*1 = 3*7*6 = 126.

Another way of looking at the same problem is just a 9C5 (or 9C4 - the two are the same): you're trying to count the number of ways to position the 5 Ds (or the 4 Ls) around the string of 9 letters (for example, 1st, 2nd, 3rd, 4th, 5th places, or 2nd, 3rd, 4th, 5th, 6th, etc.), and the remaining spaces go automatically to the Ls.

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Thu Oct 14, 2010 2:48 am

rb90 wrote:

The OA is [spoiler]20. I am getting 10 thriugh 2C1*5C4.[/spoiler]
Please explain where i am going wrong.
Thanks in advance guys.

To travel to the bus, Casey has to move down 5 times (DDDDD) and to the left 4 times (LLLL). Thus, any arrangement of the 9 letters DDDDDLLLL will yield a possible route, because it will include 5 movements downward and 4 movements to the left.

The total number of ways to arrange 9 elements is 9!. But we have to account for the repeated elements in our arrangement, the 5 D's and the 4 L's. These repeated elements will decrease the number of unique arrangements. Whenever we have a repeated element in a permutation, we have to divide by (the number of repetitions)!.

So in this problem we need to divide by 5! to account for the 5 D's and by 4! to account for the 4 L's.

Number of ways to arrange DDDDDLLLL = 9!/(5!*4!) = 126.

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rb90: Senior | Next Rank: 100 Posts; Posts: 77; Joined: Mon Jul 12, 2010 12:14 am; Location: India; GMAT Score:710

by rb90 » Thu Oct 14, 2010 2:54 am

Im so sorry. I posted the wrong quesiton by mistake. The OA to this is 126. Thanks for the help because i was as it is going to post this question as well.

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