Math - probability

S

steeze*nspice

Member
i missed this day so how do i set up the following question please. thank you.

Twelve face cardsa re removed from a deck of fifty-two cards. from the face cards, three card hands are dealt. Determine the number of distinct three card hands that are possible which include

A) no restrictions

B) 3 Kings

C) 1 Queen 2 Kings

 
A) there are no restrictions. so you can have any combination of kings, queens and jacks. there are four suits of each, and thus 12 distinct cards. any combination of 3 cards is possible, but the order does not matter. i.e. "queen of hearts, king of spades, jack of clubs" is the same as "jack of clubs, queen of hearts, kind of spades." okay?

the formula for this on your calculator will be the "C" function. 12C3 (the 12 and 3 will be subscripts) and the math goes like this:

12C3 = 12!/3!9! = (12x11x10...x2x1)/(3x2x1x9x8x7...2x1) = 12x11x10/3x2x1 = 220

B) there are 4 distinct kings in a deck, and with a three-card hand where order is not significant, you use the same function:

4C3 = 4!/3! = 4

C) there are 4 distinct queens and 4 distinct kings, so the math is a little more complicated for this one:

4 (the number of queens) x 4C2 = 4 x 4!/2! = 4 x 6 = 24

hope that helped.
 
haha shit! at least i got one of them right! good thing i cant do that, but i have my multivariable calc final tomorrow!
 
i do probability completely different to all of you, but i really cbf explaining considering the answer is already up there.

i got taught how to do it without a calculator...so it would be massive longer.
 
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