A/K vs QQ

[Cyrus]

Tekari,

The shortcut (more shortcuts! [img]/forums/images/icons/smile.gif[/img]) to the formula provided by BruceZ in his response, and the way to calculate all combos, is the following:

You want to know how many combinations you can make by picking 5 different items every time from a total of 47 items. That's C(47,5).

What you do is you put in the denominator the number 5!, or 5factorial, in other words 5*4*3*2*1. The numbers in the multiplication are of course 5.

Start the factorial of 47 in the numerator, only cut off the sequence after 5 numbers again. In other words, the numerator will be 47*46*45*44*43.

Which means that C(47,5) = (47*46*45*44*43) / (5*4*3*2*1) = 1,533,939.

To calculate C(47,2) by hand the same thing would have been done --- but I do this on Excel. [img]/forums/images/icons/smile.gif[/img]

Take care.

--Cyrus

PS : This simplification is extensively presented also in Sklansky's "Getting The Best Of It".

PPS : Note that when you are faced with a monster such as C(47,42), which asks you to get the total number of combinations with each combination having 42 items out of a total of 47 items, this is actually equal to C(47,5). That 5 is simply 47-42=5. It's much easier to calculate the latter than the former, and they give the same result!
In general, for every n>k, C(n,k)=C(n,m), whereby m=n-k.

[BruceZ]

Tekarki,

Like I already mentioned:

C(n,k) = n*(n-1)*(n-2)*...*(n-k+1) / [k*(k-1)*(k-2)*...*1]

This is the "shortcut" of "cutting off the sequence" in the numerator, defined precisely. Nothing more to the "shortcut" or of course I would have presented it. Nothing more needed other than an understanding.

-Bruce

[BruceZ]

Actually, if your calculator has a factorial key (!) but no key for C(n,k), then the above so-called shortcut is anything but. The real shortcut is n!/k!/(n-k)!. That's a lot shorter than multiplying a bunch of numbers.

[Cyrus]

I referred to "a shortcut to the formula given by BruceZ", meaning the complete formula for combinations, C(n,k)= n! / [k!*(n-k)!] , which BruceZ posted (DeGroot p.28). BruceZ went on to simplify this in mathematical terms, C(n,k) = n*(n-1)*(n-2)*...*(n-k+1) / [k*(k-1)*(k-2)*...*1]. I gave a practical step-by-step, a summary actually of what Sklansky provides in "Getting The Best Of It".

So, yes, the shortcut is as described above by the notable contributor to this forum. (Also, in the formula posted by BruceZ [img]/forums/images/icons/grin.gif[/img] )

--Cyrus

[I see that BZ has taken some kind of vow not to respond directly to my posts ( [img]/forums/images/icons/frown.gif[/img] ) or even address me by name ("certain posters"). I'll follow suit this time and respond to my own post, too. Maybe there's some method to this.]