Problem Solving

191. Pat will walk from Intersection X to Intersection Y
along a route that is confined to the square grid of
four streets and three avenues shown in the map
above. How many routes from X to Y can Pat take
that have the minimum possible length?
(A) 6
(B) 8
(C) 10
(D) 14
(E) 16

Is it possible to solve this sum by some formula ?

bhumika.k.shah wrote:191. Pat will walk from Intersection X to Intersection Y
along a route that is confined to the square grid of
four streets and three avenues shown in the map
above. How many routes from X to Y can Pat take
that have the minimum possible length?
(A) 6
(B) 8
(C) 10
(D) 14
(E) 16

Is it possible to solve this sum by some formula ?

Please provide the "Map above"

Please find the same attached.

bhumika.k.shah wrote:Please find the same attached.

The destination is 3up and 2 Left from the current place

Now its a matter of arranging 3Us and 2Ls

5!/3!*2! = 120/6*12 =10

got it till the highlighted part.

next is what???

ajith wrote:

bhumika.k.shah wrote:Please find the same attached.

The destination is 3up and 2 Left from the current place

Now its a matter of arranging 3Us and 2Ls

5!/3!*2! = 120/6*12 =10

bhumika.k.shah wrote:got it till the highlighted part.

next is what???

ajith wrote:

bhumika.k.shah wrote:Please find the same attached.

The destination is 3up and 2 Left from the current place

Now its a matter of arranging 3Us and 2Ls

5!/3!*2! = 120/6*12 =10

There is nothing left - we're done

Each of the arrangement is one solution to the problem
For example UUULL is one ULULU is another ULLUU is another .... so and so forth we have 10 solutions which takes only 5 steps to reach the destination. (ULULU means 1 unit up then 1 unit left then 1 up, then 1 left and final up)

So 10 aka C is the answer.

I am a little weak in permutation. Could you explain how u came to 5!/3!*2! = 120/6*12 =10

thnks

okletsdothis LETS DO THIS

we have to cross total 5 lines in grid to reach from X to Y. so 5!
but then we can choose 3 streets out of 4 and 2 avenues out 3 to get there. (even if you choose any way you will have to cross 3 streets and 2 avenues). so 3!.2!


hope you got it....if not we can do this again

Hi Saurabh,

thnkx for replying but i still did not understand. So i searched a bit and in one of the posts Stuart has replied.

https://www.beatthegmat.com/pat-walking- ... 49172.html

Check it out. He really takes the point hope especially with his examples.

best of luck!!


-okletsdothis !!

bhumika.k.shah wrote:got it till the highlighted part.

next is what???

ajith wrote:

bhumika.k.shah wrote:Please find the same attached.

The destination is 3up and 2 Left from the current place

Now its a matter of arranging 3Us and 2Ls

5!/3!*2! = 120/6*12 =10

Whenever an element is repeated in a permutation, we have to divide by (number of repetitions)!.

For example:

The number of ways to arrange the letters in the word SPEED is 5!/2! = 60. We divide by 2! to account for the 2 E's.

The number of ways to arrange the letters in the word RADAR is 5!/(2!*2!) = 30. We divide by 2! to account for the 2 A's and by another 2! to account for the 2 R's.

The number of ways to arrange the letters in the word MISSISSIPPI = 11!/(4!*4!*2!). We divide by 4! to account for the 4 S's, by another 4! to account for the 4 I's, and by 2! to account for the 2 P's.

In the problem above, we have to make 2 movements east and 3 movements north: EENNN. Any arrangement of the letters EENNN will yield a possible route.

Number of ways to arrange EENNN = 5!/(2!*3!) = 10.

GMATGuruNY,


thanks alot for breaking that down!