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A 16-step path is to go from (-4,-4) to (4,4) with each step increasin 31 Mar 2019, 20:15
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Difficulty:

95% (hard)

Question Stats:

6% (02:05) correct 94% (02:41) wrong based on 35 sessions

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A 16-step path is to go from (-4,-4) to (4,4) with each step increasing either the x-coordinate or the y-coordinate by 1. How many such paths stay outside or on the boundary of the square \(-2 \leq x \leq 2\), \(-2 \leq y \leq 2\) at each step?

A. 92
B. 144
C. 1,568
D. 1,698
E. 12,800
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D
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A 16-step path is to go from (-4,-4) to (4,4) with each step increasin 01 Apr 2019, 19:59
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Bunuel , chetan2u
can you give explanation for this question.
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A 16-step path is to go from (-4,-4) to (4,4) with each step increasin 01 Apr 2019, 22:51
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Ways to get from (-4,-4) to (4,4) = 16!/(8!*8!) = 12870 (A)

Ways to get from (-4,-4) to (-2,-2) = \(\frac{4!}{(2!*2!)} = 6\) (B)

Ways to get from (-2,-2) to (2,2) =\(\frac{8!}{(4!*4!)}= 70\) (C)

The actual answer to the problem ( Unable to arrive at the official answer using any of the above numbers). As per me, it should have been (A- (B*C))

(Used the method as explained in this video on Maze solution using P&C)
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A 16-step path is to go from (-4,-4) to (4,4) with each step increasin 02 Apr 2019, 06:58
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Bunuel
A 16-step path is to go from (-4,-4) to (4,4) with each step increasing either the x-coordinate or the y-coordinate by 1. How many such paths stay outside or on the boundary of the square \(-2 \leq x \leq 2\), \(-2 \leq y \leq 2\) at each step?

A. 92
B. 144
C. 1,568
D. 1,698
E. 12,800
Show more


Hi globaldesi
As Bunuel too has mentioned, it is hard and I would say it is even hard for the standards of Indian CAT as it will take some time.
But I will solve it underneath, some point may be useful in some other question.

First point - In m*n board, the ways to reach from left bottom corner to opposite corner moving up or right is \(\frac{(m+n)!}{m!n!}\)

So here we have to reach from (-4,-4) to (4,4), that is 8*8 board, so straight \(\frac{(8+8)!}{8!8!}\), but the restriction is the path cannot be inside the BLUE square. 8*8 is like a chess board, so ill take an image of chess board and work over it. The brown lines are x-axis and y-axis.

So we have to go into three cases..
(I) Case I - A to B and then B to T... A to B - \(\frac{(6+0)!}{6!0!}=1\), and B to T - \(\frac{(2+8)!}{8!2!}=45\).. Total 1845=45.
(II) Case II - A to C and then C to T... A to C - \(\frac{(6+1)!}{6!1!}=7\), but subtract one of case I, that is A to B to C, so 7-1=6 , and C to T - \(\frac{(2+7)!}{7!2!}=36\).. Total 6*36=216.
(II) Case III - A to D and then D to T... A to D - \(\frac{(6+2)!}{6!2!}=28\), but subtract 1+6 of case I and case II, that is 28-7=21 , and C to T - \(\frac{(2+6)!}{6!2!}=28\).. Total 21*28=588
Total 45+216+588=849

The above was from the left of the blue box, but similar number of ways will be from below the blue box, so another 849.
TOTAL = 2*849=1698.

D
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A 16-step path is to go from (-4,-4) to (4,4) with each step increasin 23 Sep 2025, 08:25
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been doing gmatclub 805+ fr a while and frequently see u solvin tough problems.....but dis one was really a level above tough id say.....at least it is to do under 5 minutes.....keep up the great work man.....thanks.....
chetan2u



Hi globaldesi
As Bunuel too has mentioned, it is hard and I would say it is even hard for the standards of Indian CAT as it will take some time.
But I will solve it underneath, some point may be useful in some other question.

First point - In m*n board, the ways to reach from left bottom corner to opposite corner moving up or right is \(\frac{(m+n)!}{m!n!}\)

So here we have to reach from (-4,-4) to (4,4), that is 8*8 board, so straight \(\frac{(8+8)!}{8!8!}\), but the restriction is the path cannot be inside the BLUE square. 8*8 is like a chess board, so ill take an image of chess board and work over it. The brown lines are x-axis and y-axis.

So we have to go into three cases..
(I) Case I - A to B and then B to T... A to B - \(\frac{(6+0)!}{6!0!}=1\), and B to T - \(\frac{(2+8)!}{8!2!}=45\).. Total 1845=45.
(II) Case II - A to C and then C to T... A to C - \(\frac{(6+1)!}{6!1!}=7\), but subtract one of case I, that is A to B to C, so 7-1=6 , and C to T - \(\frac{(2+7)!}{7!2!}=36\).. Total 6*36=216.
(II) Case III - A to D and then D to T... A to D - \(\frac{(6+2)!}{6!2!}=28\), but subtract 1+6 of case I and case II, that is 28-7=21 , and C to T - \(\frac{(2+6)!}{6!2!}=28\).. Total 21*28=588
Total 45+216+588=849

The above was from the left of the blue box, but similar number of ways will be from below the blue box, so another 849.
TOTAL = 2*849=1698.

D
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