Find the number of ways in which the number 30 can be partitioned into three unequal parts,each part being a natural number.What this number would be if equal parts are also included.
Let the three natural number into which 30 is partitioned are $n_1,n_2,n_3$
so $n_1+n_2+n_3=30,n_1,n_2,n_3\ge1$
I solved it to get $\binom{24+6-1}{6-1}$ but the answers given are $61,75.$Can someone explain how these answers were obtained.
On
Suppose we are distributing 30 identical balls into 3 identical jars. Suppose first the jars are distinguishable.
Then, using stars and bars counting, there are $\binom{29}{2}=29\times 14$ ways of distribution. Denote a distribution by an ordered 3-tuple. Of these, one such distribution $(10, 10, 10)$ has all three jars equal, and $13\times 3$ distributions have exactly two jars equal - i.e., a permutation of $(a, a, b)$ where $a\neq b$. Why is this? Because if we set $2a+b=30$, then the $b$ value which ranges from ${2, 4, 6, ..., 28}$ determines the $a$ value, but we must ignore $a=10$. This gives 13 possible $b$ values, with 3 permutations each.
For the first part, the number of distributions for distinct jars with all different numbers of balls is $\binom{29}{2} - 1 - 3\times13.$ To get back to indistinguishable jars, we divide this through by 3!. This is valid since we now only have distributions of the form $(a,b,c)$ with $a,b,c,$ all distinct, and so each are overcounted exactly 6 times. This should get you 61.
For the second part, add in the 1+13 options we discounted previously, corresponding to $(10, 10, 10)$ and $(a, a, b)$ with $a\neq b$.
Consider a partition of $30$ into three unequal parts $p_1 > p_2 > p_3$. Write $d_1 = p_1 - p_2$ and $d_2 = p_2 - p_3$. Then $p_1 = d_1 + d_2 + p_3$ and $p_2 = d_2 + p_3$, and hence
$$30 = d_1 + 2d_2 + 3p_3.\tag{1}$$
It's easy to see that any solution to $(1)$ in positive integers corresponds uniquely to a partition of $30$ into unequal parts. Now, counting the number of solutions ot $(1)$ in positive integers can be done easily via recursion.
If we do want to account for the possibility of equal parts, then our partitions satisfy $p_1\geqslant p_2\geqslant p_3$, and $p_1 = p_2 \iff d_1 = 0$ and $p_2 = p_3 \iff d_2 = 0$. Our problem is now like $(1)$, except we allow $d_1$ and $d_2$ to be $0$. Alternatily, with $e_1 = d_1 + 1$ and $e_2 = d_2 + 1$, we wish to solve
$$33 = e_1 + 2e_2 + 3p_3 \tag{2}$$
in the positive integers, and this is honestly not much different from what we've already done.
Let's count the number of solutions to equation $(1)$ in the positive integers, and hopefully you can do the same for the next bit. I actually find it easier to count if we allow our variables to be $0$, so let's instead write $d_1 = a_1 + 1$, $d_2 = a_2 + 1$ and $p_3 = b_3 + 1$. Hence we wish to count the solutions to
$$a_1 + 2a_2 + 3b_3 = 24$$
in the nonnegative integers.
By inspection, $b_3$ can take values in $\{0, 1, \dots, 8\}$, so any solution to $(1)$ corresponds to a solution (in the nonnegative integers) of exactly one of the equations below:
$$\begin{array}{ccc} a_1 + 2a_2 &=& 24\\ a_1 + 2a_2 &=& 21\\ a_1 + 2a_2 &=& 18\\ a_1 + 2a_2 &=& 15\\ a_1 + 2a_2 &=& 12\\ a_1 + 2a_2 &=& 9\\ a_1 + 2a_2 &=& 6\\ a_1 + 2a_2 &=& 3\\ a_1 + 2a_2 &=& 0\\ \end{array} $$
Hence, all we need to do is devise a way to count the solution to $a_1 + 2a_2 = 3k$, where $a_1$, $a_2$ and $k$ are nonnegative integers.
The value of $a_2$ uniquely determines $a_1$, so we need only consider a solution to $0\leqslant a_2 \leqslant \frac32 k$ in the positive integers. Of course, there are $\left\lfloor \frac32k\right\rfloor + 1$ such solutions. It follows that the number of solutions is given by
$$\sum_{k=0}^{8} \left(\left\lfloor \frac32 k\right\rfloor + 1\right)= 9 + \sum_{k=0}^{8} \left\lfloor \frac32 k\right\rfloor$$
which we may yet simplify based on the parity of $k$. For even $k$ we have that $\frac32 k$ is an integer, so we can do away with the floors. With $k = 2j$ this yields
$$\sum_{j=0}^4 3j = 30$$
For odd $k$, writing $k = 2j+1$, we have
$$\left\lfloor \frac32 k\right\rfloor = \left\lfloor \frac{6j+3}2 \right\rfloor = 3j + 1,$$
which yields
$$\sum_{j=0}^3 \left(3j + 1\right) = 4 + \sum_{j=0}^3 3j = 22.$$
Hence, the total number of solutions for the first part is $9+30+22 = 61$, as expected.