20. Let $a, b, c, d$ be positive integers such that the least common multiple (L.C.M.) of any three of them is $3^{3} \times 7^{5}$. How many different sets of $(a, b, c, d)$ are possible if the order of the numbers is taken into consideration?
(2 marks)
已知 $a 、 b 、 c 、 d$ 爲正整數, 當中任意三個數的最小公倍數 (L.C.M. ) 圴爲 $3^{3} \times 7^{5}$ 。若需考虑各數的先後次序, 問 $(a, b, c, d)$ 有多少組不同的可能値?
20. 11457
20. Note that each of $a, b, c, d$ is of the form $3^{m} 7^{n}$ where $m$ is one of $0,1,2,3$ and $n$ is one of 0,1 , $2,3,4,5$. Furthermore, at least two of the four $m$ 's are equal to 3 and at least two of the four $n$ 's are equal to 5 .
We first count the number of ways of choosing the four $m$ 's. There are three possibilities:
- If all four of them are 3 , there is 1 choice.
- If exactly three of them are 3 , there are 4 ways to choose the outlier and 3 ways to choose the value of the outlier (namely, 0,1 or 2 ), giving $4 \times 3=12$ choices.
- If exactly two of them are 3 , there are $C_{2}^{4}=6$ ways to locate the 3 's, then $3^{2}=9$ ways to choose the value of the other two $m$ 's, giving $6 \times 9=54$ choices.
Hence there are $1+12+54=67$ ways to choose the values of the $m$ 's. Likewise, there are $1+4 \times 5+C_{2}^{4} \times 5^{2}=171$ ways to choose the values of the $n$ 's. It follows that the answer is $67 \times 171=11457$.