# Doubt in high school permutations result of identical items

Find the total number of proper factors of $7875$

For solving this question I used a result of identical Permutations:

The total number of selections of some or all out of $p+q+r$ items where $p$ are alike of one kind, $q$ are alike of second kind and rest are alike of third kind is $[(p+1)(q+1)(r+1)]-1$.

This formula works perfectly fine for this question as $(2+1)(3+1)(1+1)-1$ = $23$

But,I can't solve question (2) by this formula

$(2)$The number of positive integral solutions of $abc= 30$ is

It's correct answer is $27$

But according to formula it's answer will be $(1+1)(1+1)(1+1)-1 = 7$

Where am I wrong?

• I had to remove your bounding box since it did not fit the space reserved for questions. – N. F. Taussig Sep 13 '18 at 17:54
• @N.F.Taussig sir please answer my question – Bobbey Ashley Sep 13 '18 at 17:58
• The number of proper factors of $30$ (which is $7$) is an entirely different question than the number of solutions for $a,b,c$ so that $abc=0$. – fleablood Sep 13 '18 at 18:15
• It's a different question. You aren't asked how many factors $30$ has, but how many factorizations of the form $abc$. Note that $2\cdot3\cdot5$ is one such factorization; $3\cdot2\cdot5$ is another. – saulspatz Sep 13 '18 at 18:16
• You're not stupid. You're learning mathematics. It's all part of the learning process. Feel good about the fact that you made progress. – David K Sep 14 '18 at 2:39

The formula you used works for determining the number of proper factors of a positive integer with three distinct prime factors. If $$n = p_1^pp_2^qp_3^r$$ then each factor of $n$ has the form $$m = p_1^{j}p_2^{k}p^{l}$$ where $j \in \{0, 1, 2, \ldots, p\}$, $k \in \{0, 1, 2, \ldots, q\}$, and $l \in \{0, 1, 2, \ldots, r\}$. Hence, $j$ can be chosen in $p + 1$ ways, $k$ can be chosen in $q + 1$ ways, and $l$ can be chosen in $r + 1$ ways. Thus, $n$ has $$(p + 1)(q + 1)(r + 1)$$ factors, of which $1$ is proper. Hence, $n$ has $$(p + 1)(q + 1)(r + 1) - 1$$ proper factors.
The first question is asking for the number of proper factors of $7875$. To determine this, we first factor $7875$ into primes. \begin{align*} 7875 & = 3 \cdot 2625\\ & = 3 \cdot 3 \cdot 875\\ & = 3 \cdot 3 \cdot 5 \cdot 175\\ & = 3 \cdot 3 \cdot 5 \cdot 5 \cdot 35\\ & = 3 \cdot 3 \cdot 5 \cdot 5 \cdot 5 \cdot 7\\ & = 3^2 \cdot 5^3 \cdot 7 \end{align*} Each factor of $7875$ has the form $3^a5^b7^c$, where $a \in \{0, 1, 2\}$, $b \in \{0, 1, 2, 3\}$, and $c \in \{0, 1\}$. Hence, there are $3 \cdot 4 \cdot 2 = 18$ factors of $7875$ and $3 \cdot 3 \cdot 2 - 1 = 17$ proper factors of $7875$.
Since $30 = 2 \cdot 3 \cdot 5$, it has $2 \cdot 2 \cdot 2 - 1 = 8 - 1 = 7$ proper factors.
The second question is asking for the number of ordered triples $(a, b, c)$ of positive integers such that $abc = 30$. This is not the same thing.
In this question, each prime must appear once among the three factors. If we write \begin{align*} a & = 2^{x_1}3^{y_1}5^{z_1}\\ b & = 2^{x_2}3^{y_2}5^{z_2}\\ c & = 2^{x_3}3^{y_3}5^{z_3} \end{align*} then \begin{align} x_1 + x_2 + x_3 & = 1 \tag{1}\\ y_1 + y_2 + y_3 & = 1 \tag{2}\\ z_1 + z_2 + z_3 & = 1 \tag{3} \end{align} since each prime appears once in the factorization of $30$. Equations 1, 2, and 3 are equations in the nonnegative integers, each of which has three solutions, depending on which of the three variables is equal to $1$. Hence, there are $3 \cdot 3 \cdot 3 = 27$ such ordered triples.