Obtain equation 5.76 by induction. The combinatorial question is this: How many different ways you can put N identical balls into d baskets (never mind the subscriptfor this problem). You could stick all of them into the third basket, or all but one in the second basket and one in the fifth, or two in the first and three in the third and all the rest in the seventh, etc. Work it out explicitly for the cases$\mathbf{N}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{N}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{N}\mathbf{=}\mathbf{3}$and$\mathbf{N}\mathbf{=}\mathbf{4}$; by that stage you should be able to deduce the general formula.

• A differential equation that uses the wave-like characteristics of particles in a field to describe matter in terms of quantum mechanics. The probability density of a particle in space and time is relevant to the response.
• The time-dependent Schrödinger equation is represented as

_{$\mathit{i}\sout{\mathbf{h}}\frac{\mathbf{d}}{\mathbf{d}\mathbf{t}}\left|\psi \left(t\right)>=\hat{H}\right|\mathit{\psi }\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{>}$}

For $N=1$, balls can be put any of d baskets

So, there are d ways.

For $N=2$

1) We can put two balls in any of d baskets ways$\Rightarrow dways$

2) We can put one ball in one of baskets, and the other in one of the remaining $d-1$

Because balls are identical, we must divide by$2\Rightarrow \frac{d(d-1)}{2}$

Total number:

$d+\frac{d(d-1)}{2}=\frac{d(d+1)}{2}$ways

1) We can put three balls in one ofbaskets$\Rightarrow d$.

2) We can put one in one basket, and two in one of remaining $d-1\Rightarrow d(d-1)$

3) We can put one ball in one of baskets, second in one of the remaining $d-1$, and third in one of remaining $d-2$.

Because balls are identical, we must divide by $3!\Rightarrow \frac{d(d-1)(d+2)}{3!}$

Total number:

$d+d(d-1)+\frac{d(d-1)(d-2)}{6}=d^2+\frac{d(d^2-3d+2)}{6}=\frac{d(d+1)(d+2)}{6}$

For $N=4$

1) We can put four balls in one of d baskets $\Rightarrow d$.

2) We can put one in one basket, and three in one of remaining$d-1\Rightarrow d(d-1)$

3) We can put two balls in one basket, and two in one of the remaining$d-1\Rightarrow \frac{d(d-1)}{2}$

4) We can put two balls in one basket, one in one of the remaining$d-1$, and last one in one of the remaining$d-2\Rightarrow \frac{d(d-1)(d-2)}{2}$

5) All four balls are in different baskets.$\Rightarrow \frac{d(d-1)(d-2)(d-3)}{4!}$

Total number of balls

$$\begin{gathered} d+\frac{d(d-1)}{2}+\frac{d(d-1)(d-2)}{6}+\frac{d(d-1)(d-2)(d-3)}{24}=\frac{d(d+1)(d+2)(d+3)}{24} \\ =d^2+\frac{d{(d-1)}^2}{2}+\frac{d(d-1)(d-2)(d-3)}{24} \\ =\frac{d}{24}\left(24d+12{(d-1)}^2+(d-1)(d-2)(d-3)\right) \\ =\frac{d}{24}\left(12d^2+12+d^3-5d^2+6d-d^2+5d-6\right) \\ =\frac{d}{24}(d^3+6d^2+11d+6) \\ =\frac{d(d+1)(d+2)(d+3)}{24} \end{gathered}$$ So, for number of balls in d_thebasket is

localid="1658227901414" $$\begin{gathered} f(N,d)=\frac{d(d+1)(d+2)+..............(d+N-1)}{N!} \\ f(N,d)=\frac{(d+N-1)!}{N!(d-1)!} \\ f(N,d)=\frac{(d+N)!}{(n+1)!(d-1)!} \end{gathered}$$

By using induction

$\mathrm{N}=0\Rightarrow \mathrm{f}(0,\mathrm{d})=1$

We assume that following expression is valid:$\mathrm{f}(\mathrm{N},\mathrm{d})=\frac{(\mathrm{d}+\mathrm{N}-1)!}{\mathrm{N}!(\mathrm{d}-1)!}$.

We check for$N+1$:

$$\begin{gathered} \left(\begin{array}{c}\mathrm{d}+\mathrm{N}\\ \mathrm{N}+1\end{array}\right)=\left(\begin{array}{c}\mathrm{d}+\mathrm{N}-1\\ \mathrm{N}\end{array}\right)+\left(\begin{array}{c}\mathrm{d}+\mathrm{N}-1\\ \mathrm{N}+1\end{array}\right) \\ =\frac{(\mathrm{d}+\mathrm{N}-1)!}{\mathrm{N}!(\mathrm{d}-1)!}+\frac{(\mathrm{d}+\mathrm{N}-1)!}{(\mathrm{N}+1)!(\mathrm{d}-2)} \\ =\frac{(\mathrm{d}+\mathrm{N}-1)!}{\mathrm{N}!(\mathrm{d}-1)(\mathrm{d}-2)!}+\frac{(\mathrm{d}+\mathrm{N}-1)!}{(\mathrm{N}+1)\mathrm{N}!(\mathrm{d}-2)!} \\ =\frac{(\mathrm{d}+\mathrm{N}-1)!}{\mathrm{N}!(\mathrm{d}-2)!}\left(\frac{1}{\mathrm{d}-1}+\frac{1}{\mathrm{N}+1}\right) \\ =\frac{(\mathrm{d}+\mathrm{N}-1)!}{\mathrm{N}!(\mathrm{d}-2)!}\frac{\mathrm{N}+\mathrm{d}}{(\mathrm{d}+1)(\mathrm{N}+1)} \\ =\frac{(\mathrm{d}+\mathrm{N})!}{(\mathrm{N}+1)!(\mathrm{d}-1)!} \end{gathered}$$

Therefore the derived expression is$(N,d)=\frac{(d+N-1)!}{N!(d-1)!}$ .