I need help with this How many strings of six hexadecimal di

I need help with this!

How many strings of six hexadecimal digits ( 0,1, 2,.......,9, A, B, ....,F) do not have any repeated digits?

Solution

Question: How many strings of six hexadecimal digits (0, 1, 2,......., 9, A, B, ....,F) do not have any repeated digits?

Description of Hexadecimal: In any base system, we have digits 0 through one-less-than-our-base. This means that, in hexadecimal, we need to have \"digits\" 0 through 15. To do this, we need single solitary digits that stand for the values of \"ten\", \"eleven\", \"twelve\", \"thirteen\", \"fourteen\", and \"fifteen\". But we don\'t. So, instead, we use letters. That is, counting in hexadecimal, the sixteen \"numerals\" are:

[ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F ]

In other words, A is \"ten\" in \"regular\" numbers, B is \"eleven\", C is \"twelve\", D is \"thirteen\", E is \"fourteen\", and \"F\" is fifteen. It is this use of letters for digits that makes complete hexadecimal numbers.

Now, here our question asks for total numbers of six hexadecimal digits that do not have repeated digits. Let us solve now as following.

Let us consider the strings of length six with characters from the set { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, A, B, C, D, E, F}

Now, we have to find out that how many strings of six hexadecimal digits do not have repeated digits. For this, the number of such strings equals the numbers of 6-permutation chosen from the set.

i.e., this number is P (16, 6) = 16! / (16-6)! = 16*15*14*13*12*11*10! / 10! = 16*15*14*13*12*11 = (16*15)*(14*13)*(12*11) = (240)*(182)*(132) = (240*182)*132 = 43680*132 = 5765760

Hence, Total 5765760 strings of six hexadecimal digits (0, 1, 2,......., 9, A, B, ....,F), do not have any repeated digits.

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