Tough question-2

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Tough question-2

by \'manpreet singh » Thu Nov 15, 2012 2:49 am
Image
[/img]

(A) 6
(B) 8
(C) 10
(D) 14
(E) 16

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by Anurag@Gurome » Thu Nov 15, 2012 2:54 am
'manpreet singh wrote:Image
[/img]

(A) 6
(B) 8
(C) 10
(D) 14
(E) 16
Image

For the length to be minimum, Pat should eight go upwards or right. So, for this he goes 3 steps up and then 2 steps right or 2 steps right and then 3 steps up, which makes 5 steps in all.
So, number of routes from X to Y that Pat can take having the minimum possible length = 5C2 = 5!/(3!2!) = 10

The correct answer is C.
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by GMATGuruNY » Thu Nov 15, 2012 3:52 am
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
Let E = traveling one block eastward and N = traveling one block northward.
To travel from X to Y by a route of minimum length, Pat must travel exactly 2 blocks eastward and exactly 3 blocks northward:
EENNN.
Any arrangement of the letters EENNN will yield a viable route.

The number of ways to arrange 5 DISTINCT elements = 5!.
But the elements above are not distinct: there are 2 identical E's and 3 identical N's.
When an arrangement includes IDENTICAL elements, we must DIVIDE by the number of ways to arrange the identical elements.
The reason is that the arrangement doesn't change when the identical elements swap positions, REDUCING the number of unique arrangements:

To illustrate:
The number of ways to arrange the letters in the word SPEED = 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 = 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 at hand:
The number of ways to arrange EENNN = 5!/(2!*3!) = 10.

The correct answer is C.

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