Discrete math. Can someone please answer this while giving step by step of how y

Question

Discrete math. Can someone please answer this while giving step by step of how you got the answer.

Consider a number system in which the k-th digit can have values 0,.....,k; so the rightmost digit is 0 or 1, the next 0,1, or 2, the next 0, 1, 2, or 3 etc. Counting in this system from one to twenty is 1 ,10, 11, 20, 21, 100, 101, 110, 111, 120, 121, 200, 201, 211, 220, 221, 300, 301, 310. The largest numbers that can be expressed with one to four digits are 1 (value 1), 21 (value 5), 321 (value 23), 4321 (value 119). Show that the value of the k-digit number represented by 100..........0 (with k - 1 zeros) is k!. Hint: The number of steps from 1 to 1000 is the same as the numbers of steps from 1001 to 2000, or from 2001 to 3000 etc; but there is no number from 5000, instead we have 10000.

Explanation / Answer

The k-digit number represented by 10000.........0 (with k-1) zeros is the smallest number with k digits.

First let us find the number of permutations of numbers with k-1 or lesser digits in the given base system.

Let us first add the number '0' to this list

The last digit can come in 2 ways. ( 0,1)

The last but one digit can come in 3 ways. (0,1,2)

The last but second digit can come in 4 ways. (0,1,2,3)

And so on upto the first digit which can be formed in k ways. (0,1.....k)

So the number of different numbers that can be formed with k-1 digits including the number 0

= 1 * 2* 3 * 4 * 5* ....k = k!

So the number of different numbers that can be formed with k-1 digits (excluding 0)

= k! - 1

Thus there are numbers in this base whose decimal value ranges from 1 to k!-1

So the greatest number in this base with k digits has the value k! -1

Thus the smallest number in the base with k+1 digits has the value k!-1+1 = k!

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