A stack of 10 different cards are shuffled and spread out face down. If 3 cards are turned face up, how many different 3-card combinations are possible?

The different 3-card combinations that are possible are; 120 possible combinations

This is a combination probability problem with the formula;

nCr = n!/((n-r)!r!)

Where;

n is sample size

r is possible choice

In this case;

n = 10

r = 3

Thus;

The different 3-card combinations possible are;

10C3 = 10!/((10 - 3)!3!)

>> 10!/(7! × 3!)

>> 120

Thus,there are 120 possible combinations of 3 different cards.

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Step-by-step explanation:

Using the combination formula as shown;

nCr = n!/(n-r)!r!

If a stack of 10 different cards are shuffled and spread out face down. If 3 cards are turned face up, the number of different ways 3cards combinations are possible is expressed as;

10C3 = 10!/(10-3)!3!

10C3 = 10!/(7)!3!

10C3 = 10*9*8*7!/7!*3*2

10C3 = 10*9*8/3*2

10C3 = 720/6

10C3 = 120 different ways

Hence there are 120 different card combinations

we have three similar triangles, because each has a right angle and shares an angle.   let's write the angles in order: opposite to short leg, long leg, hypotenuse.

cab similar to bad similar to cbd

or as ratios,

ca: ab: cb = ba: ad: bd = cb: bd: cd

we also know

ac = ad + cd

(ad+cd): ab: cb = ba: ad: bd = cb: bd: cd

(ad+cd)/cb=cb/cd

we have cb=6, ad=5 and seek x=cd.

we reject the negative root and conclude x=4

answer: 4

96 39 cause 135 -96=39 im smart (i used a calculater)