the number of terms in expansion of (a-b+c)^3 is
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Answer:
A problem of this nature can be solved by considering basic concepts from combinatorics. Lets say that we have the expansion
(a1+a2+a3+...+an)r
This implies that each of the unique terms in the expansion would have a total of ‘r’ elements which maybe repeated or not. For example, the first element would be ak1 which essentially has k number of repeated a1 's.
Now, according to combinatorics, the number of ways n objects can be combined uniquely taking r at a time is given by (n+r−1)!/r!(n−1)! . This is the general case.
For the given question, n = 4 since there are four elements a, b, c and d. r = 3 since we have a cubic power and so will form groups of 3 elements in each term (like a3,a2b and so on). So, using the formula above:
(4+3−1)!/3!(4−1)!=6!/(3!3!)=20.
Similarly, we can find for the simpler case (a+b)2:
Here, n = r = 2. So, the number of terms
3!/2!1!=3. Which is true since it evaluates to a2+b2+2ab
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