nPr is n!/(n-r)!. The ! is factorial; for example 5! = 5*4*3*2*1.
The number of permutations of a set of distinct objects is calculated using the factorial of the number of objects. For the numbers 10 through 14, there are 5 distinct numbers (10, 11, 12, 13, and 14). Therefore, the number of permutations is 5! (5 factorial), which equals 5 × 4 × 3 × 2 × 1 = 120.
The number of permutations of 8 distinct things is given by 8 factorial, denoted as 8!. This is calculated as 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1, which equals 40,320. Therefore, there are 40,320 permutations of 8 distinct items.
The number of permutations of ( n ) objects taken ( n-1 ) at a time is given by the formula ( P(n, n-1) = \frac{n!}{(n - (n-1))!} = \frac{n!}{1!} = n! ). Therefore, the number of such permutations simplifies to ( n \times (n-1)! ), which equals ( n! ). Thus, for any positive integer ( n ), the number of permutations of ( n ) objects arranged ( n-1 ) at a time is ( n \times (n-1)! ).
The letters in "PNRCSE" are all unique, and since there are six letters, the number of permutations of these letters taken six at a time is simply the factorial of 6. Thus, the number of permutations is 6! (6 factorial), which equals 720.
32 divided by 2 equals 16. 48divided by 3 equals 16. 64 divided by 4 equals 16. 16 divided by 1 equals 16.
The number of permutations of a set of distinct objects is calculated using the factorial of the number of objects. For the numbers 10 through 14, there are 5 distinct numbers (10, 11, 12, 13, and 14). Therefore, the number of permutations is 5! (5 factorial), which equals 5 × 4 × 3 × 2 × 1 = 120.
The number of permutations of 8 distinct things is given by 8 factorial, denoted as 8!. This is calculated as 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1, which equals 40,320. Therefore, there are 40,320 permutations of 8 distinct items.
The number of permutations of ( n ) objects taken ( n-1 ) at a time is given by the formula ( P(n, n-1) = \frac{n!}{(n - (n-1))!} = \frac{n!}{1!} = n! ). Therefore, the number of such permutations simplifies to ( n \times (n-1)! ), which equals ( n! ). Thus, for any positive integer ( n ), the number of permutations of ( n ) objects arranged ( n-1 ) at a time is ( n \times (n-1)! ).
31
Mass divided by volume equals density. For the same volume, if the mass is more then the density is higher.
If 6a divided by 2 equals 12 then A equals
32 divided by 2 equals 16. 48divided by 3 equals 16. 64 divided by 4 equals 16. 16 divided by 1 equals 16.
No. 35 divided by 8 equals 4.375 .35 divided by 5 equals 7 .
1 divided by 13 equals 13
30 divided by 2 equals 15 :)
3.5111
6