"Combination count" Fseq

Title Description

answer

Now our problem into the next figure from the origin, the following two actions:

• Plus abscissa $1$ , the ordinate plus $1$ , this case corresponds to a selected number of $1.$
• Plus abscissa $1$ , the ordinate Save $1$ , this case corresponds to a selected number of $-1.$

How many kinds of programs without $y = -1$ , reach $(n+m,n-m).$

There is a $n+m$ operations, a total of m times can be selected for operation down, so the answer is: $C_{n+m}^{m}$.

Now consider after $y = -1$ answer, this time from $(0,0)$ to $y = -1$ can be called from $(0,-2)$ to $y = -1$ , and therefore can be converted from line $(0,-2)$ to $(n+m,n-m)$ the number of programs. compared to $(0,0)$ to $(n+m,n-m)$ , the more opportunity to rise, the less the chance of a fall, so the answer is: $C_{n+m}^{m-1}$.

So in the end we are asking the probability should be: $\frac{C_{n+m}^{m}-C_{n+m}^{m-1}}{C_{m}^{n+m}}=1-\frac{m}{n+1}$

Then the conclusion is a problem, and expressed difficulty.