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.