Yes. The total number of combinations (i.e. tickets) picking k sized combinations from a pool of n events is denoted by the following math shorthand:
Code:
n
# of combinations = ( k )
To calculate the above, use the following formula:
# of combinations = n! / (k! * (n-k)!)
Th math shorthand "n!" denotes "n factorial". it equals the product of all the positive integers less than or equal to n. So, in the case of n = 10, 10! = 10*9*8*7*6*5*4*3*2*1 = 3,628,800.
As you can see, factorials can become very large values, so some shortcuts are handy at calculating the # of combinations given (n,k).
In your example n = 10, and k = 4. So, the formula becomes:
# of combinations = 10!/(4! * (10-4)!) = 10!/(4! * 6!)
To perform this calculation quickly, note that 10! can also be expressed as 6!*7*8*9*10. So, when dividing 10! by 6!, the terms for 6! cancel out, leaving 10*9*8*7.
So the formula can now be expressed:
# of combinations = 10*9*8*7/(4*3*2*1)
Note that 2, 3, and 4 in the denominator divide respective terms in the numerator, that is 8/4, 9/3, and 10/2. The 7 term does not divide so it remains. The resulting formula is now:
# of combinations = 10/2 * 9/3 * 8/4 * 7 (note I dropped the 1 term as it does not change any value in multiplication)
and then:
# of combinations = 5 * 3 * 2 * 7
which equals 210.
Also, you should observe that if k = 6, you get the same # of combinations, 210.
Last edited by PokerHero77; 06-28-2022 at 08:22 PM.
