+363 Let $n$ be a positive integer.
(a) There are $n^2$ ordered pairs $(a,b)$ of positive integers, where $1 < a,$ $b < n.$ Using a counting argument, show that this number is also equal to n+2(n2).
(b) There are $n^3$ ordered triples $(a,b,c)$ of positive integers, where $1 < a,$ $b,$ $c < n.$ Using a counting argument, show that this number is also equal to n+3n(n−1)+6(n3).
+1950 Alright, I'll try to tackle this problem!
(a) We can try to count the number of ordered pairs (a, b) ourselves. We can split these into two cases.
1. a=b
2. a≠b
1. Since we can choose any numbers between 1 and n, there are obviously n possibilities we can achieve.
2. Since we can choose any 2 numbers between 1 and n that are not equal, we just have (n2).
Hence, we just have n+(n2) +(n2), which is exactly what we wanted in the first place.
I hope I answered part (a) correctly!
Thanks! :)
+1950 (b) For question b, we basically do the same exact thing as stated above.
First, let's review the cases for a=b=c.
There are obviously n such triplets.
Next, let's look at when exactly two are the same. You have 3 choices for which number is repeated and then n choices for the repeated number, and then n-1 for the distinct number. This gets us 3n(n-1).
Lastly, we have 3 different numbers. There are 6(n3) ways because there are 3 distinct numbers out of n and 3! = 6 ways to do this.
Thus, when we add them together, we get n+3n(n−1)+6(n3).!
Thanks! :)
