Find the coefficient of x^3Y^2z^2 in (x+y+z)^7
( a + b + c) ^ n = Ó n ! / ( i! j! k! ) a ^ i b ^ j c ^ k
where i + j + k = n
So exponent of x ^ 3 y ^ 2 z ^ 2 =
7! / (3! 2! 2!) =
5040 / ( 6 * 2 * 2 ) =
5040 / 24 =
210
Ó mean summ
To find the coefficient of x^3Y^2z^2 in the expansion of (x+y+z)^7, we can use the Binomial Theorem. The Binomial Theorem states that the expansion of (a + b)^n can be written as:
(a + b)^n = C(n, 0) * a^n * b^0 + C(n, 1) * a^(n-1) * b^1 + C(n, 2) * a^(n-2) * b^2 + ... + C(n, n) * a^0 * b^n
Where C(n, k) represents the binomial coefficient, which is the number of ways to choose k items from a set of n items.
In our case, we are looking for the coefficient of x^3Y^2z^2, so we want to find the term in the expansion that has x^3Y^2z^2.
We can use the formula for the binomial coefficient to calculate the coefficient. The binomial coefficient C(n, k) is given by the formula:
C(n, k) = n! / (k!(n-k)!), where n! represents n factorial.
Let's substitute the values into the formula:
C(7, 3) = 7! / (3!(7-3)!)
Simplifying:
C(7, 3) = 7! / (3! * 4!)
= (7 * 6 * 5 * 4 * 3 * 2 * 1) / ((3 * 2 * 1) * (4 * 3 * 2 * 1))
= 35
So, the coefficient of x^3Y^2z^2 in (x+y+z)^7 is 35.