## In a certain game, a one-inch square piece is placed in the

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### In a certain game, a one-inch square piece is placed in the

by varun289 » Fri Apr 05, 2013 8:45 pm
In a certain game, a one-inch square piece is placed in the lower left corner of an eight-by-eight grid made up of one-inch squares. If the piece can move one grid up or to the right, what is the probability that the center of the piece will be exactly inches away from where it started after 8 moves?

64/256 , 70/256 ?

which one is ans?

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by misterholmes » Sat Apr 06, 2013 9:55 am
Exactly how many inches away?
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### Re: In a certain game, a one-inch square piece is placed in the

by GMATGuruNY » Fri Sep 11, 2020 12:35 pm
In a certain game, a one-inch square piece is placed in the lower left corner of an eight-by-eight grid made up of one-inch squares. If the piece can move one grid up or to the right, what is the probability that the center of the piece will be exactly 4√2 inches away from where it started after 8 moves?

A. 24/256

B. 64/256

C. 70/256

D. 128/256

E. 176/256 The figure above shows a 4 by 4 dotted square with a diagonal of 4√2.
Question stem, rephrased:
If the blue dot starts at X, moves 8 times, and can move only eastward or northward, what is the probability that the blue dot finishes at Y?

Let E = moving one square eastward and N = moving one square northward.

Good outcomes:
To travel from X to Y in 8 moves, the blue dot must move exactly 4 squares eastward and exactly 4 squares northward:
EEEENNNN.
Any arrangement of the letters EEEENNNN will yield a viable route.

The number of ways to arrange 8 DISTINCT elements = 8!.
But the elements above are not distinct: there are 4 identical E's and 4 identical N's.
When an arrangement includes IDENTICAL elements, we must DIVIDE by the number of ways each set of identical elements can be arranged.
The reason:
The arrangement doesn't change when the identical elements swap positions, REDUCING the number of unique arrangements.
Here, we must divide by 4! to account for the four identical E's and by another 4! to account for the four identical N's:
8!/(4!4!) = 70

All possible outcomes:
Since there are 2 options for each move -- E or N -- and 8 moves in total, we get:
2*2*2*2*2*2*2*2 = 256

Thus:
P = (good outcomes)/(all possible outcomes) = 70/256