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

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


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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

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SquareGrid.JPG [ 25.84 KiB | Viewed 2618 times ]
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C

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] New post   24 Nov 2018, 23:43
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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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Re: In a certain game, a one-inch square piece is placed in the lower left [#permalink] 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] 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.
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Re: In a certain game, a one-inch square piece is placed in the lower left [#permalink] New post   25 Nov 2018, 20:16
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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] New post   27 Aug 2020, 05:40
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Given

    • 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.
    • the piece can move one grid up or to the right.


To Find

    • The probability that the center of the piece will be exactly 4√2 inches away from where it started after 8 moves.

Approach and Working Out

    • Total number of moves = \(2^8\)
      o = 256

    • To be 4\(\sqrt{2 }\) inches away it must have 4 upward and 4 rightward movement.
      o That can be done in 8C4 ways = 70 ways.

    • Probability = 70/256

Correct Answer: Option B
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In a certain game, a one-inch square piece is placed in the lower left [#permalink] New post   27 Aug 2020, 06:09
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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




The figure above shows a 4 by 4 dotted square with a diagonal of \(4\sqrt{2}\).
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:
\(\frac{8!}{4!4!} = 70\)

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:
\(P = \frac{good-outcomes}{all-possible-outcomes} = \frac{70}{256}\)

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Re: In a certain game, a one-inch square piece is placed in the lower left [#permalink] New post   21 Jul 2023, 19:17
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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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