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

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In a certain game, a one-inch square piece is placed in the lower left  [#permalink]

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New post Updated on: 25 Nov 2018, 11:33
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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 \(4\sqrt{2}\) inches away from where it started after 8 moves?

A. 24/256

B. 64/256

C. 70/256

D. 128/256

E. 176/256


I can understand total sample space is 256 i.e. \(2^8\). I am not able to figure out why there are 70 possible ways. As per the explanation provided :

"
Since there are eight moves, and there are two choices each move, there are 2^8 moves. Now figure out how many will move the piece exactly inches away. 4\sqrt{2} is the diagonal formed by a square four inches by four inches; thus a five-by-five grid. So, you need to move four moves up and four moves to the right. Next, figure out how many ways we could get four “up” moves: 8*7*6*5. There are eight places where the first “up” could go, seven where the next “up” and so on. //Since you don’t care which “up” is which, you have to get rid of the duplicates: (8*7*6*5)/ (4*3*2*1). So the probability is 70/256 "

As it was mentioned that the square piece is placed at the lower left corner, I can think of only 6 ways. Please refer to the attached diagram. I have marked the probable path as 1,2,..6. The Boxes mentioned as 3 are the boxes those were traversed when the 3rd path is followed.

In the below post I can see 7C4 is mentioned. I am aware of the concept of C and P. If someone wishes to explain by this method, please feel free.

[url]https://gmatclub.com/forum/in-a-certain-game-a-one-inch-square-piece-is-placed-in-the-64302.html
[/url]

Another suggestion required, how to make the options as radio buttons.

Regards,
Arup

Attachment:
SquareGrid.JPG
SquareGrid.JPG [ 25.84 KiB | Viewed 477 times ]

Originally posted by ArupRS on 24 Nov 2018, 07:12.
Last edited by Bunuel on 25 Nov 2018, 11:33, edited 2 times in total.
Renamed the topic and edited the question.
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Re: In a certain game, a one-inch square piece is placed in the lower left  [#permalink]

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New post 24 Nov 2018, 23:43
1
ArupRS wrote:
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\sqrt{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




Now basically it means there are 8 steps either vertical or horizontal, so what we have in the end is a straight line or different right angled triangles as shown in the attached figure..
so the distance between initial and final position as marked by star is \(4\sqrt{2}\), so we are looking at right angled triangle otherwise a straight line will b eof 8 units
Only way the hypotenuse can be \(4\sqrt{2}\) is when the the right angled triangle is isoeceles as A and B are integer units and A+B=8.
thus we move 4 steps horizontally and 4 vertically..
ways to move
Choose 4 steps out of 8 that can be moved horizontally and rest 4 automatically take care of 4 vertical steps, so 8C4=\(\frac{8!}{4!4!}=70\)
Total ways to move 8 steps
Each step would be horizontal or vertical, thus total ways = \(2^8\)

thus probability = \(\frac{70}{2^8}\)

C
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star.png
star.png [ 30.2 KiB | Viewed 423 times ]


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Re: In a certain game, a one-inch square piece is placed in the lower left  [#permalink]

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New post 25 Nov 2018, 00:13
UUUURRRR = 8!/4!4!
Total space= 2^8

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Re: In a certain game, a one-inch square piece is placed in the lower left  [#permalink]

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New post 25 Nov 2018, 11:31
chetan2u wrote:
ArupRS wrote:
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\sqrt{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




Now basically it means there are 8 steps either vertical or horizontal, so what we have in the end is a straight line or different right angled triangles as shown in the attached figure..
so the distance between initial and final position as marked by star is \(4\sqrt{2}\), so we are looking at right angled triangle otherwise a straight line will b eof 8 units
Only way the hypotenuse can be \(4\sqrt{2}\) is when the the right angled triangle is isoeceles as A and B are integer units and A+B=8.
thus we move 4 steps horizontally and 4 vertically..
ways to move
Choose 4 steps out of 8 that can be moved horizontally and rest 4 automatically take care of 4 vertical steps, so 8C4=\(\frac{8!}{4!4!}=70\)
Total ways to move 8 steps
Each step would be horizontal or vertical, thus total ways = \(2^8\)

thus probability = \(\frac{70}{2^8}\)

C


chetan2u : I am still having hard time to understand how come we can have more than 6 steps, after we fixed the position in lower left corner. Could you please add one path other than those I have highlighted in the attached image. Doing so may help me to understnad other options which I am missing now.
Attachments

SquareGrid.JPG
SquareGrid.JPG [ 25.84 KiB | Viewed 349 times ]

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Posts: 7764
Re: In a certain game, a one-inch square piece is placed in the lower left  [#permalink]

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New post 25 Nov 2018, 20:16
1
ArupRS wrote:
chetan2u wrote:
ArupRS wrote:
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\sqrt{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




Now basically it means there are 8 steps either vertical or horizontal, so what we have in the end is a straight line or different right angled triangles as shown in the attached figure..
so the distance between initial and final position as marked by star is \(4\sqrt{2}\), so we are looking at right angled triangle otherwise a straight line will b eof 8 units
Only way the hypotenuse can be \(4\sqrt{2}\) is when the the right angled triangle is isoeceles as A and B are integer units and A+B=8.
thus we move 4 steps horizontally and 4 vertically..
ways to move
Choose 4 steps out of 8 that can be moved horizontally and rest 4 automatically take care of 4 vertical steps, so 8C4=\(\frac{8!}{4!4!}=70\)
Total ways to move 8 steps
Each step would be horizontal or vertical, thus total ways = \(2^8\)

thus probability = \(\frac{70}{2^8}\)

C


chetan2u : I am still having hard time to understand how come we can have more than 6 steps, after we fixed the position in lower left corner. Could you please add one path other than those I have highlighted in the attached image. Doing so may help me to understnad other options which I am missing now.


Hi..

You are missing out on all zig-zag options..
One-two example..
_, _, _, _, F
_, _, 7, 7, 7
_, _,7, _, _
7, 7, 7,_, _
S, _, _, _, _,

So start-up -right-right-up-up-right-right-F

Many more similar ways
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Re: In a certain game, a one-inch square piece is placed in the lower left   [#permalink] 25 Nov 2018, 20:16
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