Combinatronics, MGMAT
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- Geva@EconomistGMAT
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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.
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.
- GMATGuruNY
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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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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
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