Question :
What is the number of ordered triplets of natural numbers $a, b, c$ such that $a b c=2^5 \times 3^7$ and H.C.F. $(a, b, c)=1$
My approach :
I tried making cases where the exponent of $2$ is "shared" between two numbers and the last one is $3^7$, then I did the same the other way round, then one having 2, one with 3, and one with both.
It was really tedious and I got the wrong answer $180$ [ and the text book itself had the wrong answer $432$ ] whereas the right answer is $315$.
On
Pick two to share 2 then 5 identical balls 2 distinct boxes $$ {}^3 \mathrm{C}_2 {}^{5+2-1} \mathrm{C}_{2-1}=18 $$
Now we do the same for 3 $$ {}^3 \mathrm{C}_2 {}^{7+2-1} \mathrm{C}_{2-1}=24 $$
So the total required is $18×24=432$
The error with your solution is that you assumed only the last number could have 3
On
The answer $432$ is not correct. The correct answer $315$ was given by Perm but later deleted because of unclear reasons. I will repeat here his arguments.
The problem is equivalent to the problem of distributing $5$ black balls and $7$ white balls between three bins, provided that balls of the same color cannot be put in all three bins.
Without the latter restriction the balls can be distributed in $$ \binom{5+2}{2}=21\quad\text{and}\quad\binom{7+2}{2}=36 $$ ways, respectively (see stars and bars).
From this we should subtract the number of cases where all three bins contain at least one ball of the given color. To find the number we place one ball of the color in each bin and count the number of ways to distribute the rest ($2$ and $4$, respectively) balls between the bins. One obtains: $$ \binom{2+2}{2}=6\quad\text{and}\quad\binom{4+2}{2}=15 $$ such distributions, respectively, so that the final answer to your question is: $$ (21-6)\times(36-15)=15\times21=315. $$
On
The gcd(a,b,c) is 1, so one is not divisible by 2, and one is not divisible by 3. There are 15 ways to distribute the powers of 2, 21 ways to distribute the powers of 3, so the total is 315.
(The 15 are 0,n,5-n for 1 <= n <= 4, times 3 because each of a,b,c can be 0, and 0,0,5 times 3).
On
This problem can also be treated well with PIE the principle of inclusion-exclusion. This answer contains a fairly detailed explanation. If you are familiar with PIE, you might only want to look at the section marked with the bar.
We consider a set $E$ of properties \begin{align*} \color{blue}{E=\{e_1,e_2\}} \end{align*} Here
$\color{blue}{e_1}$ is the property, that $2$ is a common factor of $a,b$ and $c$, i.e. $\color{blue}{2|\gcd(a,b,c)}$
$\color{blue}{e_2}$ is the property, that $3$ is a common factor of $a,b$ and $c$, i.e. $\color{blue}{3|\gcd(a,b,c)}$.
We define for sets $T\subseteq E$: \begin{align*} \color{blue}{N_{\supseteq T}} &=\#\{(a,b,c)\in\mathbb{N}^3:abc=2^53^7, \gcd(a,b,c) \mathrm{\ possesses\ }\bf{at\ least}\\ &\qquad\qquad\qquad\qquad\quad\mathrm{\ the\ factors\ according\ to\ properties\ }\ e_j\mathrm{\ in\ }\it{T}\}\\ \color{blue}{N_{= T}} &=\#\{(a,b,c)\in\mathbb{N}^3:abc=2^53^7, \gcd(a,b,c) \mathrm{\ possesses\ }\bf{exactly}\\ &\qquad\qquad\qquad\qquad\quad\mathrm{\ the\ factors\ according\ to\ properties\ }\ e_j\mathrm{\ in\ }\it{T}\}\\ \end{align*}
Since we are looking for the numbers of triples $(a,b,c)$ with $abc=2^53^7$ with \begin{align*} \gcd(a,b,c)=1 \end{align*} we want to calculate $\color{blue}{N_{=\emptyset}}$ which means that neither $2$ nor $3$ is a divisor of $\gcd(a,b,c)$ and consequently $\gcd(a,b,c)=1$.
According to PIE we calculate \begin{align*} \color{blue}{N_{=\emptyset}} &\color{blue}{=N_{\supseteq\emptyset}-N_{\supseteq \{e_1\}}-N_{\supseteq \{e_2\}}+N_{\supseteq \{e_1,e_2\}}}\tag{1}\\ &=\binom{7}{2}\binom{9}{2}-\binom{4}{2}\binom{9}{2} -\binom{7}{2}\binom{6}{2}+\binom{4}{2}\binom{6}{2}\tag{2}\\ &=21\cdot 36-6\cdot 36-21\cdot 15+6\cdot 15\\ &\,\,\color{blue}{=315} \end{align*} according to the claim.
Comment:
In (1) we use PIE and calculate $N_{=\emptyset}$ the number of triples $(a,b,c)$ with $abc=2^53^7$ and $\gcd(a,b,c)=1$ as
$\color{blue}{N_{\supseteq\emptyset}:}$ the number of all triples $(a,b,c)$ with $abc=2^53^7$
$\color{blue}{-N_{\supseteq \{e_1\}}:}$ minus the number of triples $(a,b,c)$ which have the factor $2$ (or more factors) dividing $\gcd(a,b,c)$
$\color{blue}{-N_{\supseteq \{e_2\}}:}$ minus the number of triples $(a,b,c)$ which have the factor $3$ (or more factors) dividing $\gcd(a,b,c)$
$\color{blue}{+ N_{\supseteq \{e1,e_2\}}:}$ plus the number of triples $(a,b,c)$ which have the factor $2$ and $3$ dividing $\gcd(a,b,c)$, which is a compensation for subtracting these numbers twice when calculating $-N_{\supseteq \{e_1\}}-N_{\supseteq \{e_2\}}$.
In (2) we use stars and bars to calculate the numbers $N_{\supseteq T}$ with $T\subseteq E$ from the right-hand side of (1). Let's for instance have a closer look at \begin{align*} \color{blue}{N_{\supseteq \{e_1\}}} \end{align*} We consider $abc=2^53^7$ where according to property $e_1$ we have that $2|\gcd(a,b,c)$. This implies we have $2|a, 2|b$ and $2|c$ and there are two factors $2$ left and seven factors $3$ left for distributing them with zero or more factors to $a, b$ and $c$ respectively.
Two factors $2$ distributed to three variables $a,b$ and $c$ gives $\binom{2+3-1}{3-1}=\binom{4}{2}$
Seven factors $3$ distributed to three variables $a,b,$ and $c$ gives $\binom{3+7-1}{7-1}=\binom{6}{2}$
Since these two distributions do not depend on each other the total number is $\binom{4}{2}\binom{6}{2}$.
The calculation of the other numbers $N_{\supseteq\emptyset}, N_{\supseteq \{e_2\}}$ and $N_{\supseteq \{e_1,e_2\}}$ follows accordingly.
PROBLEM :
Let us say we have $3$ "Buckets" where we want to "Distribute" $5$ "Black" Balls & $7$ "White" Balls.
Let us also say that all $3$ Buckets having a Common Colour Ball is not allowed.
It is "almost" Equivalent to what you want.
CALCULATION :
Total ways of Distributing $5$ Black Balls : Partitioning $5$ into $3$ Parts.
There is a formula to get that , though we can avoid that formula in this simple case.
We can see that :
$(0,a,5-a)$ will work for $a=0,\cdots,5$ : 6 ways
$(1,a,4-a)$ will work for $a=0,\cdots,4$ : 5 ways
$(2,a,3-a)$ will work for $a=0,\cdots,3$ : 4 ways
$(3,a,2-a)$ will work for $a=0,\cdots,2$ : 3 ways
$(4,a,1-a)$ will work for $a=0,\cdots,1$ : 2 ways
$(5,a,0-a)$ will work for $a=0,\cdots,0$ : 1 ways
Total $6 \times 7/2 = 21$ ways
Like-wise for the White Balls , we have $8 \times 9/2 = 36$
Total ways $21 \times 36 = 756$
It is "almost" Equivalent to the Number Case.
We are over-counting the Cases where Common Colour Ball might occur in all Buckets. We want to eliminate that.
Put $1$ Black Ball each in the $3$ Buckets.
We have $2$ Balls , which we can "Partition" like $(2,0,0),(0,2,0),(0,0,2),(1,1,0),(1,0,1),(0,1,1)$ : $3 \times 4/2 = 6$ Ways are not "allowed".
Hence $21-6 = 15$ ways are "allowed".
Like-wise for White Balls , put $1$ White Ball each in the $3$ Buckets.
We have $4$ Balls , which gives $5 \times 6/2 = 15$ ways which are not allowed.
Hence "allowed" ways are $36-15 = 21$
Hence the total ways we want is $15 \times 21 = 315$
SUMMARY :
Total ways (with or without Common Coluor Ball) is $756$
Total ways without Common Colour Ball is $315$ (this is what we want)
$756-315=441$ ways are not allowed due to Common Colour Ball.
ADDENDUM :
I changed my line of attack : Instead of checking what was wrong here , I decided to list the various ways & figure out what was wrong with the other answer. That will show what is wrong where. I was able to figure the Core Issue !
Here are the $6$ ways to Distribute the $5$ Black Balls in 2 Buckets , leaving the first Bucket Empty :
$\color{red}{(0,0,5)},(0,1,4),(0,2,3),(0,3,2),(0,4,1),\color{red}{(0,5,0)}$
Now , the other answer multiplies this by $3$ to get $3 \times 6 = 18$ , because the Empty Bucket can be the middle or the last.
$\color{red}{(0,0,5)},(1,0,4),(2,0,3),(3,0,2),(4,0,1),\color{red}{(5,0,0)}$
$\color{red}{(0,5,0)},(1,4,0),(2,3,0),(3,2,0),(4,1,0),\color{red}{(5,0,0)}$
We can easily see that the first & the last (in red) will be over-counted $6$ times , though it should be counted only $3$ times , excluding the repeated elements.
Thus is it only $15$ ways.
Similarly , the other answer over-counts by $3$ to get $24$ , rather than $21$.
Hence Correct total is not $18 \times 24 = 432$ , it is $15 \times 21 = 315$
The other Answer is wrong , even though it is accepted.
WHEW !!