GMAC 11th Edition Review Guide PS Question

Question

In the 11th Edition, there is a PS problem that I cannot figure out:

Pat will walk from intersection X to intersection Y and to get there, he must walk upwards (U) twice and rightwards (R) three times. How many routes can he take to have the minimum possible length?

The answer is Ten, but I have no clue how to do it more easily. The answer lists the possible combinations but does not provide any formula.

UURRR
URRUR
URRRU
etc...

Does anyone know a faster way to figure this problem out without listing all possibilities and counting them? Thanks!

mrsmarthi · Answer

Well, this is somewhat a disguise question of P & C(Permutations and Combinations).

Patt must walk 2 ways upwards(U) and 3 ways to the right(R). In how many ways can he take the route.

This is nothing but to find out "In how ways can we arrange the letters UURRR? "

Ans - 5 ! / (2! * 3!) = 10.

jenyang5268 · Answer

Why is it that the answer to the following question is solved in the exact same way?

-How many different groups of 3 people can be formed from a group of 5 people?

Answer: 10

To solve this problem, don't I also use the equation: 5!/(3! x 2!) = 10 ? In this case, n!/k!(n-k)!.

Now I'm confused why in the previous question, there are 2U's and 3 R's, making a total of 5 letters. However, I am not selecting only 3 letters for my combination so why does is this problem solved with the formula n!/k!(n-k)! ? Please advise. Thanks!

nookway · Answer

I will take a stab at explaining this.

1. The original problem you had asked is about Permutation with repeating elements. The formula to solve those is Total number of permutations = N!/(K!*L!) where N is the number of items where K of the N elements are of the same type as well as L of the N elements are of the same type.

In your example: N = 5 (U,U,R,R,R)
K = 2 (U,U)
L=3 (R,R,R)

N!/(K!*L!) = 5!/(2!*3!) = 10

2. Removing second solution as I was thinking permutation and not combination

I hope this helps.

mrsmarthi · Answer

The two questions you are asking are different.

-How many different groups of 3 people can be formed from a group of 5 people?

This is a combination since the order in which 3 people are selected is NOT important. It is just the selection. Hence 5C3 is correct.

Applying the formula, nCk = n! / (r! * n-r!), we have 5! / (2! * 3!) = 10.

Now coming to the original question, there are 5 paths where 2 paths are upwards(U) and 3 paths are sideways(S).

To reach from source to distination, the order of the path is important. So this means it is Permutation.

Let me take an example - How many ways can we arrange PAGES. 5 distinct letters are there and order is important so it is 5!. 
How many ways can we arrange FEELS - Again 5 words with a E repeated twice. It will be 5! / 2!.

Similary coming to our problem, UUEEE can be arranged in 5! / (2! * 3!) since U is repeated twice and E is repeated thrice.

https://regentsprep.org/regents/math/permut/LpermRep.htm

Google for Permutations with Repitions and you should find some information.

jenyang5268 · Answer

Oh ok great! Thanks a lot =)

GMAC 11th Edition Review Guide PS Question

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