Question

Let $n$ be a positive integer. (a) Prove that \[n^3 = n + 3n(n - 1) + 6 \binom{n}{3}\]by counting the number of ordered triples $(a,b,c)$ of positive integers, where $1 \le a,$ $b,$ $c \le n,$ in two different ways. (b) Prove that \[\binom{n + 2}{3} = (1)(n) + (2)(n - 1) + (3)(n - 2) + \dots + (k)(n - k + 1) + \dots + (n)(1),\]by counting the number of subsets of $\{1, 2, 3, \dots, n + 2\}$ containing three different numbers in two different ways.

Answer 1

Answer:

Description of the given points can be defined as follows:

Step-by-step explanation:

In points A, If b is less than a and c. so, it calls the smiley face (a, b , c), so we'll have a smiling smile whenever we create a chart! And if b is bigger then a and c, then it should be an angry face, although when we switch the head inside out, they get a grumpy face.    

In point b,  the smirks, which including a greater than b and c, then it is the neutral faces like a is equal to and b equal to c , it can also be seen in certain forms of faces. When the facial is positive, therefore b and b is similar to c. because when we pick a, b and c, but there's a choice of n are neutral.

Its quantity of scowls has been counted. There are many n options to choose a or n-1 forms of picking b. Each value that we can choose for the could also have been the price of b or c. There is also multiplication by 3. Thus 3n(n-1) scowls were present.  

A number of happy face faces we note now. There will be n options for such a then n – 1 options for b, but n – 2 options for c. There were the happy faces of n(n-1)(n-2) = 3C(n,3). There are many grumpy face faces of the configuration of 3C(n,3).  

Thus, n^3 = n + 3n(n – 1)+ 6C(n,3) was its total amount of faces.

(b) We can select two groups of 2 categories one at two numbers to choose three numbers and the other with one group. Its total number of n+2 can be divided into two groups.  

First, there are many two numbers with one group, then n number in another one. This makes up n + 2 number of all. C(2,2) = 1 represents selecting two digits from the two sets. There are many C(n,1) = n forms to select a single organization amount. Which offers us a first 1*n term.