16 number it have
That would be: 20! / 10! = 20 * 19 * 18 * 17 * 16 * 15 * 14 * 13 * 12 * 11 = 670442572800 * * * * * No! That is the number of permutations! The number of combinations is 20C10 = 20!/(10!*10!) = 20*19*18*17*16*15*14*13*12*11/(10*9*8*7*6*5*4*3*2*1) = 184,756
120 four letter permutations if you don't allow more than one 'o' in the four letterarrangement.209 four letter permutations if you allow two, three and all four 'o'.1.- Let set A = {t,l,r,m,}, and set B = {o,o,o,o}.2.- From set A, the number of 4 letter permutations is 4P4 = 24.3.- 3 letters from set A give 4P3 = 24, and one 'o' can take 4 different positions in theword. That gives us 24x4 = 96 four letter permutations.4.- In total, 24 + 96 = 120 different four letter permutations.5.- If the other three 'o' are allowed to play, then you have 2 letters from set A thatgive 4P2 = 12 permutations and two 'o' can take 4C2 = 6 position's, giving 12x6 = 72four letter permutations.6.- One letter from set A we have 4P1 = 4, each one can take 4 different positions, therest of the spaces taken by three 'o' gives 4x4 = 16 different permutations.7.- The four 'o' make only one permutation.8.- So now we get 72 + 16 + 1 = 89 more arrangements adding to a total of 89 + 120 = 209 different 4 letter arrangements made from the letters of the word toolroom.[ nCr = n!/((n-r)!∙r!); nPr = n!/(n-r)! ]
16
A credit card number is a 16 digit number.
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).
There are 216 permutations of three dice. Of these, 206 have a sum that is less than 16, specifically, the permutations 466, 556, 565, 566, 646, 655, 656, 664, 665, and 666 have sums that are 16 or greater - all other permutations have sums that are less than 16. The probability, then, of rolling a sum less than 16 on three dice is 206 in 216, or about 0.9537.
Rethink the question as "How many ways can the letters of MISSISSIPI (one P) be arranged?" The (initial) answer is that this is the number of permutations of 10 things taken 10 at a time, because every time you choose a P, you simply write down two P's. That is 10 factorial, which is 3,628,800. However, since there are still multiple letters (four S's, and four I's), you need to divide by 24 twice in order to see how many distinct permutations there are. That is 3,628,800 / 16 / 16, or 14,175.
If order matters and a number cannot be repeated, then there are 16x15x14x13 = 43680 different combinations. If order does not matter then you must divide by the number of ways you can rearrage the same 4 numbers i.e. divide by 4x3x2x1=24 so you can select 4 numbers from a group of 16 in 43680/24 = 1820 ways. In mathematical terms there are 43680 permutations of 16 numbers taken 4 at a time and 1820 combinations of 16 number taken 4 at a time.
Base ten isn't used because it doesnt utilise all the permutations of 0's and 1's of a 4 bits, or any other number of bits. The number ten would be represented as 1010 and then you'ld have to start again at 0000. Hexadecimal utilises all 16 of the permutations between 0000 and 1111.
16 number it have
That would be: 20! / 10! = 20 * 19 * 18 * 17 * 16 * 15 * 14 * 13 * 12 * 11 = 670442572800 * * * * * No! That is the number of permutations! The number of combinations is 20C10 = 20!/(10!*10!) = 20*19*18*17*16*15*14*13*12*11/(10*9*8*7*6*5*4*3*2*1) = 184,756
120 four letter permutations if you don't allow more than one 'o' in the four letterarrangement.209 four letter permutations if you allow two, three and all four 'o'.1.- Let set A = {t,l,r,m,}, and set B = {o,o,o,o}.2.- From set A, the number of 4 letter permutations is 4P4 = 24.3.- 3 letters from set A give 4P3 = 24, and one 'o' can take 4 different positions in theword. That gives us 24x4 = 96 four letter permutations.4.- In total, 24 + 96 = 120 different four letter permutations.5.- If the other three 'o' are allowed to play, then you have 2 letters from set A thatgive 4P2 = 12 permutations and two 'o' can take 4C2 = 6 position's, giving 12x6 = 72four letter permutations.6.- One letter from set A we have 4P1 = 4, each one can take 4 different positions, therest of the spaces taken by three 'o' gives 4x4 = 16 different permutations.7.- The four 'o' make only one permutation.8.- So now we get 72 + 16 + 1 = 89 more arrangements adding to a total of 89 + 120 = 209 different 4 letter arrangements made from the letters of the word toolroom.[ nCr = n!/((n-r)!∙r!); nPr = n!/(n-r)! ]
16
A number 16 scoop measures 2 ounces. The number 16 indicates that there are 16 scoops per quart.
That depends how many are taken for each permutation. If you take 1 number at a time, then there are 20 permutations. 2 at a time . . . . . (20 x 19) = 380 3 at a time . . . . . (20 x 19 x 18) = 6,840 4 at a time . . . . . (20 x 19 x 18 x 17) = 116,280 5 at a time . . . . . (20 x 19 x 18 x 17 x 16) = 1,860,480 etc.
16