Respuesta :
There are 17 possible choices for the first ball, 16 possible choices for the second and so on.. until 6 balls are chosen. Multiply the number of choices together.
17×16×15×14×13×12 = 8,910,720
This is also what's called a permutation, 17 choose 6
In permutations order matters so (1,2,3) is not the same as (3,2,1).
17×16×15×14×13×12 = 8,910,720
This is also what's called a permutation, 17 choose 6
In permutations order matters so (1,2,3) is not the same as (3,2,1).
The number of possible outcomes are there for this game is 8,910,720
How to find total number of outcomes if a part of distinguishable items are chosen?
Suppose there are n items, all distinct from each other. You have to chose k items out of n ( [tex]k \leq n[/tex] ).
Then total number of ways this can be done is: [tex]^nP_k[/tex] where P denotes permutations(we use permutations when ordering of items matter, which does in this case when objects are distinguishable, so AB and BA are two things, for example.). This is evaluated as:
[tex]^nP_k = \dfrac{n!}{(n-k)!}[/tex]
where '!' is sign of factorial(defined only for whole numbers) such that:
[tex]0! = 0\\\forall \: n \in \mathbb Z^+; n! = n \times (n-1) \times ... \times 2 \times 1[/tex]
[tex]\mathbb Z^+[/tex] is set of positive integers.
Thus, as there are n = 17 balls which are numbered (which means they can be differentiated from one another), and k = 6 balls have to be chosen. Thus, total number of ways this can be done will the number of all possible outcomes of this game.
Thus, number of all possible outcomes of this game is evaluated as:
[tex]^nP_k = \dfrac{n!}{ (n-k)!} = n \times (n-1) \times ... \times (n-k + 1)\\\\^{17}P_{6} = \dfrac{17!}{(17-6)!} = \dfrac{17!}{11!} \\\\\\^{17}P_{6} =17 \times 16 \times 15 \times 14 \times 13 \times 12 = 8910720[/tex]
Thus, the number of possible outcomes are there for this game is 8,910,720
Learn more about permutations and combinations here:
https://brainly.com/question/11958814
