Draw 5 card, how many combination that result contains at least 1 king.?

From a common deck of 52 cards containing 4 kings, how many combination of drawing 5 cards and the result contains at least 1 king.

Or generally, from any deck of k cards containing n specific kind of cards, how many combination of drawing r cards and the result contains at least 1 card of that kind.

I have attempted to find the formula, but I am not sure that it is correct.

My solution is: summation from [i=1] to [min(n, r)] of [nCi]

∫ sin^4x cos²x dx

rearrange it as:

∫ sin²x (sinx cosx)² dx =

recall the half-angle identity:

sin²x = (1/2)[1 - cos(2x)]

also, recall the double-angle identity:

sin(2x) = 2sinx cosx → sinx cosx = (1/2) sin(2x)

so that the integral becomes:

∫ (1/2)[1 - cos(2x)] [(1/2) sin(2x)]² dx =

∫ (1/2)[1 - cos(2x)] (1/4) sin²(2x) dx =

pulling out the constants and expanding the integrand,

(1/8) ∫ [sin²(2x) - cos(2x) sin²(2x)] dx =

break it into:

(1/8) ∫ sin²(2x) dx - (1/8) ∫ sin²(2x) cos(2x) dx =

according to the above mentioned half-angle identity rewrite the first integrand as:

(1/8) ∫ (1/2){1 - cos[2(2x)]} dx - (1/8) ∫ sin²(2x) cos(2x) dx =

(1/16) ∫ dx - (1/16) ∫ cos(4x) dx - (1/8) ∫ sin²(2x) cos(2x) dx =

(1/16)x - (1/16) [(1/4) sin(4x)] - (1/8) ∫ sin²(2x) cos(2x) dx =

divide and multiply the remaining integral by 2 so as to let the derivative of sin(2x) come out:

(1/16)x - (1/64) sin(4x) - (1/8)(1/2) ∫ sin²(2x) 2cos(2x) dx =

(1/16)x - (1/64) sin(4x) - (1/16) ∫ sin²(2x) d[sin(2x)] =

(1/16)x - (1/64) sin(4x) - (1/16) [1/(2+1)][sin(2x)]^(2+1) + c =

(1/16)x - (1/64) sin(4x) - (1/16) (1/3) sin³(2x) + c

in conclusion:

∫ sin^4x cos²x dx = (1/16)x - (1/64) sin(4x) - (1/48) sin³(2x) + c



I hope it helps