It is 6!
The number of diagonals in an n-sided polygon is given by nC2 - n (where n is the number of sides of the polygon) or in the expanded form: factorial (n) _______________________ {factorial (2) * factorial (n-2)} substituting (n = 6) for a hexagon we get the number of diagonals as 9. Similarly, substituting (n=5) for a pentagon we get the number of diagonals as 5.
Well, darling, the word 'planets' has 7 letters, so you can form 7! (7 factorial) = 5040 different words. Listing them all out would take longer than a Golden Girls marathon, so I'll spare you the agony. Just trust me, there are a lot of words you can make from 'planets'.
I suspect that you want permutations rather than combinations. The permutation ABC is different from ACB, but they are both the same combination. There are 26*26*26*10*10 or 1,757,600 possible permutations of 3 letters followed by 2 numbers. But there are ten ways of arranging 3 letters and 2 numbers: eg LLLNN, LNLNL etc. All in all, therefore, 17,576,000 permutations. However, some letters are not used so as to avoid confusion between letters and numbers: eg 0 and O. Also, some sequences are not used because they form (or suggest) inappropriate words.
There are a total of 15 letters in "season greetings." To calculate the number of words that can be formed, we first need to determine the number of unique arrangements of these letters. This can be calculated using the formula for permutations of a multiset, which is 15! / (2! * 2! * 2! * 2! * 2! * 2! * 1!). This results in 1,816,214,400 unique arrangements. However, not all of these arrangements will form valid English words, as many will be nonsensical combinations of letters.
word form means when you have a number and you write the number in words instead of letters
4.5.6.7 = 7!/3!
do you mean n factorial, n! perhaps.
A mixture of letters and a number that you jumbled together to form a question!
To determine the number of permutations for forming two teams of 3 from a group of 6 people, first choose 3 people for the first team. This can be done in ( \binom{6}{3} = 20 ) ways. Since the order of teams matters, there are ( 20 \times 2 = 40 ) permutations. Therefore, there are 40 different ways to form two teams of 3 from 6 people.
report zbharath. data:num type i value 5, fac type i value 0. perform fact using num changing fac. write:/ 'factorial of',num,'is',fac. form fact. using value(f-num) type i. changing f-fact type i. f-fact=1. while f-num ge 1. f-fact=f-fact*f-num. f-num=f-num-1. endwhile. endform.
Those are the letters used for roman numbers, but written in that order, they don't form a valid number.
standard form is when your write the number in letters so it is twelve and six tenths