To find the number of 2-number combinations that can be made with the numbers 1 through 6, we use the combination formula ( C(n, r) = \frac{n!}{r!(n-r)!} ), where ( n ) is the total number of items to choose from and ( r ) is the number of items to choose. Here, ( n = 6 ) and ( r = 2 ). Thus, the number of combinations is ( C(6, 2) = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15 ). Therefore, there are 15 different 2-number combinations that can be made with the numbers 1 to 6.
16
The question does not make sense. 21/16 is one number: there is no second number in which it needs to go.
16 times
The number of arrays you can make with the number 16 depends on how you define "arrays." If you're referring to the factors of 16, they are 1, 2, 4, 8, and 16, which can form rectangular arrays of various dimensions (e.g., 1x16, 2x8, 4x4). In terms of combinations or arrangements of the number 16 in an array (like in permutations), the possibilities would be significantly greater, depending on the context and constraints you apply.
24 =16, including the null combination.
Assuming that "number" means digits and that it is permutations (rather than combinations) that are required, the answer is 816 which is 281.475 trillion (approx).
Small words make a difference. No prime number "has" a factor of 16. But 2 is the only prime number that "is" a factor of 16.
16 number it have
25 - 9 = 16
To calculate the number of doubles and trebles from 16 selections, you can use combinations. For doubles, the number of ways to choose 2 selections from 16 is given by the combination formula ( \binom{n}{r} ), which is ( \binom{16}{2} = \frac{16!}{2!(16-2)!} = 120 ). For trebles, the number of ways to choose 3 selections from 16 is ( \binom{16}{3} = \frac{16!}{3!(16-3)!} = 560 ). Thus, there are 120 doubles and 560 trebles from 16 selections.
16 x 16
Divide by 16 (the number of ounces in a pound) equals One pound (avdp).