Question

How many strings of 20-decimal digits are there that contain two 0s, four 1s, three 2s, one 3, two 4s, three 5s, two 7s, and three 9s?

Short answer

The total number of strings are 58,663,725,120,000 strings of 20-decimal digits

Step-by-step solution

Step 1: Use the formula for string factorial

Total number of arrangements of n objects if al are different = n!

If r! of them are same number of ways are given by

\(\begin{array}{l}\frac{{n!}}{{{n_1}!{n_2}!..{n_k}!}} = \frac{{n!}}{{r!}}\\r! = {n_1}!{n_2}!..{n_k}!\end{array}\)

The n! Is string length

r! are outcome of strings

Step 2: Solution of strings of 20-decimal digits

Let’s, applied r! and n! values,

Because if the element that is repeated is exchanged with the similar element at some other position, it does not create a new outcome.

In our case we have two 0s, four 1s, three 2s, one 3s, two 4s, three 5s, two 7s, three 9s and the string length is 20.

Here,

\(\begin{array}{l}n! = 20!\\r! = 2!,4!,3!,1!,2!,3!,2!,3!\\\frac{{n!}}{{r!}} = \frac{{20!}}{{2!.4!.3!.1!.2!.3!.2!.3!}}\\\frac{{n!}}{{r!}} = 58,663,725,120,000{\rm{ }}strings\end{array}\)