The five women can be arranged in a row in $5!$ ways. This creates six spaces where we can place the men, four between successive women and two at the ends of the row.
$$\square w \square w \square w \square w \square w \square$$
To separate the men, we must choose four of these spaces in which to place a man, which can be done in $\binom{6}{4}$ ways. The men can be arranged in the chosen spaces in $4!$ ways. Hence, there are
$$5!\binom{6}{4}4!$$
arrangements in which the men are separated.
Imagine there are $3$ distinct items: $A,B,C$.
(i) The number of ways of selecting one or more items from $3$ distinct items is:
$${3\choose 1}+{3\choose 2}+{3\choose 3}=2^3-1=7 \ \ \ (\text{note:} \sum_{k=0}^n {n\choose k}=2^n).$$
Indeed, the outcomes are:
$$A,B,C,AB,AC,BC,ABC.$$
Now imagine there are $3$ identical items: $A,A,A$.
(iv) The number of dividing $3$ identical items among $2$ persons, each of them who can receive 0,1,2, or more items is:
$${3+2-1\choose 2-1}={4\choose 1}=4 \ \ \ (\text{explained below})$$
Indeed, the outcomes are:
$$\{AAA,0\}, \{AA,A\}, \{A,AA\}, \{0,AAA\}.$$
Explanation: let $x_1,x_2$ be the number of identical items the two persons receive, respectively. Then, the equation is:
$$x_1+x_2=3, \ \ \ \ 0\le x_1,x_2\le 3.$$
1-method: Stars and bars. Consider the number of stars as number of items to be given to a person and a bar to switch from one person to another person. For example, several examples:
$$\begin{align}**|* &\ \ (\text{the first person gets $2$, the second gets $1$})\\
|*** &\ \ (\text{the first person gets $0$, the second gets $3$})\\
*|** &\ \ (\text{the first person gets $1$, the second gets $2$})\\
***| &\ \ (\text{the first person gets $3$, the second gets $0$})
\end{align}$$
Note that it is basically a combination of $4$ symbols (items) taken $1$ (bar) or $3$ (stars) at a time:
$${4\choose 1}={4\choose 3}=4.$$
2-method: Generating functions. The equation to be solved is:
$$x_1+x_2=3, \ \ \ \ 0\le x_1,x_2\le 3.$$
Let the equation
$$(x^0+x^1+x^2+x^3)(x^0+x^1+x^2+x^3)=x^{3}$$
represent the number of items (indicated on the exponents) the two persons (indicated by the brackets) can receive. For example:
$$x^0\cdot x^3=x^3 \ \ (\text{the first person gets $0$, the second gets $3$})\\
x^2\cdot x^1=x^3 \ \ (\text{the first person gets $2$, the second gets $1$})\\
x^3\cdot x^0=x^3 \ \ (\text{the first person gets $3$, the second gets $0$})\\
x^1\cdot x^2=x^3 \ \ (\text{the first person gets $1$, the second gets $2$})\\$$
Now if we consider the sum of $x^3$ we get $4x^3$. So, the problem reduces to finding the coefficient of $x^3$ in the expansion of the left-hand side:
$$\begin{align}[x^3](1+x+x^2+x^3)^2&=[x^3]\left(\frac{1-x^4}{1-x}\right)^2= \quad \quad \quad \quad \quad \quad \quad \quad \quad (1)\\
&=[x^3](1-x^4)^2(1-x)^{-2}= \ \quad \quad \quad \quad \quad \quad (2)\\
&=[x^3]\sum_{k=0}^2 {2\choose k}(-x^4)^k\cdot \sum_{k=0}^{\infty} {-2\choose k}(-x)^k= \ \ (3)\\
&=[x^3]{2\choose 0}(-x^4)^0\cdot {-2\choose 3}(-x)^3= \quad \quad \quad \quad (4)\\
&={2+3-1\choose 3}= \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad (5)\\
&={4\choose 3}.\end{align}$$
where:
(1) inside brackets: the sum of four terms of GeomProg.
(2) simple algebra.
(3) negative binomial series.
(4) considering only the terms, whose exponents add up to $3$.
(5) negative binomial theorem.
Best Answer
The answer provided by you is wrong. The count of $5^6$ selects and arranges 6 ice-creams in a order after selecting from 5 different ice-creams. But we want to get no of ways by which the child can buy 6 ice-creams and not to arrange them.
The correct approach can be as follows as the total no of ice creams of each type are unknown .
Let no of ice-creams bought of each type as $a,b,c,d,e$ .So, $$a+b+c+d+e=6$$ as boy needs 6 ice-creams. And thus answer is $10_{C_4}$ . Which is same as arranging 6A's and 4B's in a row.