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In a certain game, a oneinch square piece is placed in the lower left
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Updated on: 25 Nov 2018, 10:33
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In a certain game, a oneinch square piece is placed in the lower left corner of an eightbyeight grid made up of oneinch 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 fivebyfive 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/inacertaingameaoneinchsquarepieceisplacedinthe64302.html [/url] Another suggestion required, how to make the options as radio buttons. Regards, Arup Attachment:
SquareGrid.JPG [ 25.84 KiB  Viewed 347 times ]
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Originally posted by ArupRS on 24 Nov 2018, 06:12.
Last edited by Bunuel on 25 Nov 2018, 10:33, edited 2 times in total.
Renamed the topic and edited the question.



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Re: In a certain game, a oneinch square piece is placed in the lower left
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24 Nov 2018, 22:43
ArupRS wrote: In a certain game, a oneinch square piece is placed in the lower left corner of an eightbyeight grid made up of oneinch 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 stepsEach step would be horizontal or vertical, thus total ways = \(2^8\) thus probability = \(\frac{70}{2^8}\) C
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1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html
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Re: In a certain game, a oneinch square piece is placed in the lower left
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24 Nov 2018, 23:13
UUUURRRR = 8!/4!4! Total space= 2^8 Posted from my mobile device
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Re: In a certain game, a oneinch square piece is placed in the lower left
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25 Nov 2018, 10:31
chetan2u wrote: ArupRS wrote: In a certain game, a oneinch square piece is placed in the lower left corner of an eightbyeight grid made up of oneinch 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 stepsEach 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 oneinch square piece is placed in the lower left
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25 Nov 2018, 19:16
ArupRS wrote: chetan2u wrote: ArupRS wrote: In a certain game, a oneinch square piece is placed in the lower left corner of an eightbyeight grid made up of oneinch 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 stepsEach 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 zigzag options.. Onetwo example.. _, _, _, _, F _, _, 7, 7, 7 _, _,7, _, _ 7, 7, 7,_, _ S, _, _, _, _, So startup rightrightupuprightrightF Many more similar ways
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1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html
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Re: In a certain game, a oneinch square piece is placed in the lower left &nbs
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