Question

Give a combinatorial proof of the following equation:

n³ = n(n-1)(n-2) + 3n(n-1)+n
Hint: Count lists of length 3 with elements from {1, 2, ..., n}; consider lists with 3 distinct elements, lists with 2 same elements and 1 different element, and lists with 3 same elements.

Answer 1

A combinatorial proof for the equation n³ = n(n-1)(n-2) + 3n(n-1) + n counts the length-3 lists from a set {1, 2, ..., n} by considering lists with three distinct elements, two identical and one different, and three identical elements. Adding the count for each type yields the total, proving the equation.

Step-by-step explanation:

To provide a combinatorial proof for the equation n³ = n(n-1)(n-2) + 3n(n-1) + n, we can count the number of length-3 lists that can be formed from the set {1, 2, ..., n}. This set will include lists with three distinct elements, those with two identical elements and one different element, and those with three identical elements.

First, consider lists with three distinct elements; we have n choices for the first element, (n-1) choices for the second, and (n-2) choices for the third, resulting in a total of n(n-1)(n-2) unique lists.

Next, the number of lists with two identical elements and one different element can be formed as follows: n choices for the identical elements and (n-1) choices for the different element. However, since the two identical elements can be arranged in 3 different positions, we multiply by 3, giving us 3n(n-1) such lists.

Finally, there are n lists with three identical elements, one for each element in our original set.

Adding these cases together, the total number of lists is n(n-1)(n-2) + 3n(n-1) + n, which is equal to n³, as required by the original equation.