Turns out COMBINATIONS, the number of possible groups which can be formed, is the key to simplifying binomial expansion. First, a review of factorials, permutations and combinations is in order, no pun intended.
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So what is the purpose of the above analysis? It's to find patterns to eliminate computation required to expand a binomial. Find and verify a pattern and decrease or eliminate long, messy computation.
Rather than listing then counting the number of ways a list or group can occur, as shown above, factorial, permutations, and combinations facititate counting.
Factorial computes the number of ways a list of n things can be made and is the function or rule which must be used three times to express the desired COMBINATION.
The permutation of n things taken r at a time shortens the list to only r long and computes that number of ways.
The combination of n things taken r at a time removes duplicates, since order doesn't count in a group, and computes that number of ways.
Now revisit a problem completed above, using factorials and the formulas for permutation and combination.
Three factorial, 3!, computes the number of ways a list of 3 things can be made. There are 3 choices for the 1st in the list, 2 choices for the second in the list, and 1 choice for 3rd in the list: 3! = 3x2x1 = 6.
"N factorial," n!, computes the number of ways a list of n things can be made. There are n choices for the 1st in the list, n-1 choices for the second in the list, n-2 choice for 3rd in the list, so on till 3 choices, 2 choices, and 1 choice: n! = (n)(n-1)(n-2)(n-3)...(3)(2)(1).
In order to limit the number in the list to just r things, division by (n-r)! is needed. This is what the permutation does. Notice that n! is (n)(n-1)(n-2)…(n-r+1)(n-r)! and division by (n-r)! does limit the list as needed.
The permutation of 3 things taken 2 at a time, 3P2, computes the number of ways a list of 3 things can be shortened to a list of 2 things and the number of ways then computed.
Next, it is necessary to remove duplicates. Combinations does this. It takes the permutation and divides by r! to eliminate duplicates making the count of lists a count of groups.
The combination of 3 things taken 2 at a time, 3C2, removes duplicates from the list and counts the number of groups of 2 that can be made from a list of 3 things.
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