The number of 4-number selections from 19 different numbers can be found using the concept of combinations. Since the order matters in this case, it is a permutation problem.
To find the number of permutations, we can calculate using the formula:
P(n, r) = n! / (n-r)!
Here, n is the total number of options (19), and r is the number of choices we need to make (4).
So, P(19, 4) = 19! / (19-4)!
= 19! / 15!
= (19 * 18 * 17 * 16 * 15!) / 15!
= 19 * 18 * 17 * 16
Therefore, there are 19 * 18 * 17 * 16 = 65,536 possible 4-number selections.
To find the number of distinguishable ways the given letters can be arranged, we can use the concept of permutations.
Since there are repeating letters (A, B, C, D), we need to consider their frequencies.
The given letters can be arranged in the following way:
A A A B B B B C C D D D D
The number of arrangements is given by the formula:
n! / (a! * b! * c! * ...)
Here, n is the total number of letters (13), and a, b, c, ... denote the frequencies of each letter.
So, the number of arrangements = 13! / (3! * 4! * 2! * 4!)
= (13 * 12 * 11 * 10!) / (3! * 2!)
= (13 * 12 * 11) / 3!
Therefore, the number of arrangements is (13 * 12 * 11) / 3! = 286.
In a small northern city lottery, you must select four numbers from nineteen different numbers to win the big prize. The numbers have to be in a particular order. How many 4-number selections are possible? In how many distinguishable ways can the following letters be arranged? A A A B B B B C C D D D D Number of arrangements =
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