Find the coefficient of x^5y^7 in the expansion of (5x+2y)^12
12 (x)^5 (2y)^7
3
My answer which i got wrong is 220x^5(128y^7)= 28160x^5y^7
12C5 (5x)^5 (2y)^7
12C5 * 5^5 * 2^7
792 * 3125 * 128 = 316800000
Where did 792 come from ? Also, thank you for helping me
huh? You surely must see from Scott's lines of text that
792 = 12C5
To find the coefficient of x^5y^7 in the expansion of (5x+2y)^12, you can use the binomial theorem.
The binomial theorem states that for any positive integer n, the coefficient of x^k y^(n-k) in the expansion of (a + b)^n is given by the formula:
C(n, k) * a^(n-k) * b^k
where C(n, k) is the binomial coefficient, which can be calculated as:
C(n, k) = n! / (k! * (n-k)!)
In our case, n = 12, k = 5, a = 5x, and b = 2y.
Now, let's calculate the coefficient of x^5y^7 using the binomial theorem:
C(12, 5) * (5x)^(12-5) * (2y)^5
C(12, 5) = 12! / (5! * (12-5)!)
= 12! / (5! * 7!)
= (12 * 11 * 10 * 9 * 8) / (5 * 4 * 3 * 2 * 1)
= 792
(5x)^(12-5) = (5x)^7
= 5^7 * x^7
= 78125x^7
(2y)^5 = 2^5 * y^5
= 32y^5
Now, substitute these values back into the formula:
792 * 78125x^7 * 32y^5
Finally, simplify:
792 * 78125 * 32 * x^7 * y^5
= 19906560000x^7y^5
Therefore, the coefficient of x^5y^7 in the expansion of (5x+2y)^12 is 19,906,560,000 (which can also be written as 1.990656 × 10^10).