A probability problem by Mihir Mistry

Probability Level 4

On a normal chess board as shown, 'I1' and 'I2' are two ants which starts moving towards each other. Each ant moves with a constant speed. Insect 'I1' can move only to the right or upward along the lines while 'I2' can move only to the left or downwards along the lines of the chess board. What is the total number of ways the two ants can meet at same point during their trip?


The answer is 12870.

Mihir Mistry by Mihir Mistry , 24, India

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3 solutions

Ivan Koswara
Jan 11, 2017

Reverse the path of the second ant. Now the two paths together form a path from I1 to I2 where you only move up or right. This is a standard problem; 8 steps up and 8 steps right gives ( 8 + 8 8 ) = 12870 \binom{8+8}{8} = \boxed{12870} paths.

6 Helpful 0 Interesting 1 Brilliant 1 Confused
Bk Lim
Apr 5, 2015

The only place they can meet is the 9 points along the top-left to bottom-right diagonal. It takes 8 steps for both of them. [ ( 8 0 ) 2 + ( 8 1 ) 2 + ( 8 2 ) 2 + ( 8 3 ) 2 ] × 2 + ( 8 4 ) 2 = 12870 \left[\left(\begin{array}{c}8\\ 0\end{array}\right)^{2}+\left(\begin{array}{c}8\\ 1\end{array}\right)^{2}+\left(\begin{array}{c}8\\ 2\end{array}\right)^{2}+\left(\begin{array}{c}8\\ 3\end{array}\right)^{2}\right]\times2+\left(\begin{array}{c}8\\ 4\end{array}\right)^{2}=12870

4 Helpful 0 Interesting 1 Brilliant 0 Confused
Anmol Agarwal
Oct 15, 2017

Using concept of relative motion, the top insect would have to cover (8,8) distance . Therefore ,arranging VVVVVVVV HHHHHHHH

where v=vertical step and h = horizontal step . The arrangement can be done in 16C8 ways ie 12870.

3 Helpful 0 Interesting 1 Brilliant 0 Confused

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