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C
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46% (02:48) correct
54%
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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
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
SquareGrid.JPG [ 25.84 KiB | Viewed 3396 times ]
"
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 [ 25.84 KiB | Viewed 3396 times ]
24 Nov 2018, 23:43
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ArupRS
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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25 Nov 2018, 00:13
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UUUURRRR = 8!/4!4!
Total space= 2^8
Posted from my mobile device
Total space= 2^8
Posted from my mobile device
