Question:
Let $N$ be the set of all integral solutions of the equation $x_1x_2x_3x_4 = 770$. Find $N$
So, for this question I used cases. Using prime factorization we know that the factors are $7 , 2 ,5 ,11$.
Case 1:When $x_1,x_2,x_3,x_4$ are $7 , 11 , 5 ,2$ in all permutations then number of ways of arranging :$4!$
Case 2:When two of the numbers are already multiplied. Ex. $7,11,10,1$ and here first I have to choose two numbers and then arrange so:${4 \choose 2} \cdot 4!$
Case 3: When a pair of two numbers are multiplied. Ex. $77 , 10 , 1 , 1$ and here I have to choose two and then arrange:${4 \choose 2} \cdot \frac{4!}{2!}$
Case 4: When three numbers are multiplied. Ex.$7 , 110 , 1,1$ and here I have to choose three numbers then arrange:${4 \choose 3} \cdot \frac{4!}{2!}$
Case 5: When one number is $770$ and the others are $1$ then ways of arranging is $4$
Thus total no of ways ($N$) $= 292$. However the answer is given as $256$. Which case have I missed and is there a better method to approach this question. Any help will be appreciated!
Best Answer
Your case 3 is not correct. $4 \choose 2$ is the number of ways to combine two prime factors, but you count $77,10,1,1$ twice, once when you choose $7,11$ for the two and once when you choose $2,5$. This divides the cases by $2$. You must have added wrong, because that correction reduces the total and the answer comes out $256$ as desired.