Question

Is the identity $(\frac{n}{k})=(\frac{n-1}{k})+(\frac{n-1}{k-1})$ true or false. .( Enter $1$ if true or $0$ if false)

Answer 1

We have to choose k balls from a set containing n balls, hence the answers $nk$ .On the other hand, the blue ball may or may not be among the selected k balls. If the blue ball is selected, then, in fact we have chosen $k-1$ red balls from $n-1$ red balls and this can be done in $n-1k-1$ ways. If the blue ball is not selected,then we have chosen k red balls from $n-1$ ones. This can be done in $n-1k-1$ whichleads to a total of $(\frac{n-1}{k})+(\frac{n-1}{k-1})$ possibilities to choose the $k$ balls.Observation Using similar arguments we can obtain a more general identity. Let us count in how many ways one can choose k balls from a set containing n red andm blue balls. Discarding the color of the balls, the answer is $(\frac{n+m}{k})$ If we take intoconsideration the fact that the chosen k balls can be: all red, or k-1 red and 1 blue,or k-2 red and 2 blue, etc. we obtain the identity $(\frac{n+m}{k})=(\frac{n}{k})(\frac{m}{0})+(\frac{n}{k-1})(\frac{m}{1})+(\frac{n}{k-2})(\frac{m}{2})+\cdots +(\frac{n}{0})(\frac{m}{k}).$