Question

A person takes a step forward or backward, then number of ways the person takes total of 7 steps, but moves just one step away from initial position is ______

Answer 1

Given :

Total 7 steps,

Have to stop at one step away from initial position.

To find :

Number of ways to move according to given condition.

Solution:

For a 1dimensional walk along the x-axis with 2n−1 steps (odd steps), where he can only move +1 or −1 each step .

With an odd number of steps he will end in an odd position.

For ending at, position (2m−1), he must take (n+m−1) forward steps and (n−m) backward steps (such that the sum of number of steps is 2n−1 and the difference is 2m−1)).

The number of ways of doing this is according to combination relation,

$\bf = \binom{2n - 1} { n + m - 1} =^aC_r = \frac{a!}{(a-r)!r!}$

where a = total elements

r = choosen elements

So, out of 2n-1 steps , we will choose only n+m-1 steps

so,

For placing himself at one step away

he can be 1 step behind or next to the starting place,

so, Total ways can be,

So values of 2m-1 can be +1 and -1.

So values of m = 1 and 0.

It is given that total number of steps (2n-1) = 7,

so n = 4.

Putting the values of n and m,

For +1 position, n = 4, m =1

Number of ways:

$= \binom{(2 \times 4) - 1}{4 + 1 - 1} \\ = \binom{7}{4} = 35$

For -1 position, n = 4, m = -1

Number of ways :

$= \binom{(2 \times 4) - ( - 1)}{4 + 1 - ( - 1)} \\ = \binom{9}{6} = 84$

So,

Total number of ways = 35 + 84 = 119