Assuming you want a rectangular array, 1 x 18, 2 x 9, or 3 x 6.
You can only do multiplation so a whole number is divisble by another whole number when the quotient is a whole number and the remainder is 0
Think of the chairs as arrays. The dimensions of the arrays give you the factors of 18.
18 Chairs into equal rows - 6 x 3 2 x 9 18 x 1
You can have: 1 row of 36 2 rows of 18 3 rows of 12 4 rows of 9 or 6 rows of 6, so in total there are 5 ways.
There are 19 ways to do this.
You can only do multiplation so a whole number is divisble by another whole number when the quotient is a whole number and the remainder is 0
If you are Canadian (no joke), there are 618(101559956668416) ways. If you are american, there are 518(3814697265652) ways. Hope this helps!
Think of the chairs as arrays. The dimensions of the arrays give you the factors of 18.
18 Chairs into equal rows - 6 x 3 2 x 9 18 x 1
> 6.40237371 × 1015Actually, since there are four i's and two o's, the number of distinct permutations of the letters in "oversimplification" is 18!/(4!2!) = 133,382,785,536,000.
You can have: 1 row of 36 2 rows of 18 3 rows of 12 4 rows of 9 or 6 rows of 6, so in total there are 5 ways.
To determine how many ways a teacher can arrange 5 students in the front row from a total of 23 students, we use permutations since the order matters. The number of ways to choose and arrange 5 students from 23 is given by the formula ( P(n, k) = \frac{n!}{(n-k)!} ), where ( n ) is the total number of students and ( k ) is the number of students to arrange. Thus, it is calculated as ( P(23, 5) = \frac{23!}{(23-5)!} = \frac{23!}{18!} = 23 \times 22 \times 21 \times 20 \times 19 = 2,598,960 ). Therefore, there are 2,598,960 different ways to arrange 5 students in the front row.
18
The number of ways is 18C5 = 18!/(5!*13!) = 8,568 ways.
To arrange arrays with the number 18, you can consider various combinations of factors that multiply to 18. The factors of 18 are 1, 2, 3, 6, 9, and 18, allowing for arrangements such as (1, 18), (2, 9), (3, 6), and their permutations. Additionally, you can create multi-dimensional arrays, such as 2x9 or 3x6, which represent different structures for organizing the number. Overall, the arrangements depend on how you choose to group and order the factors of 18.
if order of beads matters than you get 18P12= 8.892185702*1012. if order doesn't matter than it's 18C12= 18564 ways to arrange the beads.
There are 19 ways to do this.