In a lottery daily game, a player picks four numbers from 0 to 9 (without repetition). How many different choices does the player havea) If order matters? b) If order does not matter?
Question
In a lottery daily game, a player picks four numbers from 0 to 9 (without repetition). How many different choices does the player havea) If order matters? b) If order does not matter?
Solution
a) If order matters:
The first number can be any of the 10 (from 0 to 9), so there are 10 possibilities. For the second number, since repetition is not allowed, there are only 9 possibilities left. For the third number, there are 8 possibilities left, and for the fourth number, there are 7 possibilities left.
So, the total number of choices is 1098*7 = 5040.
b) If order does not matter:
In this case, we are looking at combinations rather than permutations. The formula for combinations is nCr = n! / r!(n-r)!, where n is the total number of options, r is the number of options chosen, and '!' denotes factorial.
Here, n = 10 (numbers 0 to 9) and r = 4 (four numbers are chosen).
So, the total number of choices is 10C4 = 10! / 4!(10-4)! = 210.
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