Question

In how many ways can the digits in the number 9,164,461 be arranged?

Answer 1

First count the number of digits present in the given number. Even if the digits are repeating then count them a different number of times. Now, use the formula that n number of different things can be arranged in $n!$ ways. Now, divide this total number of ways of arrangement with the product of factorials of the number of times each digit is repeated to get the total number of effective arrangements.

Complete step-by-step solution:
Here we are provided with the number 9,164,461 and we are asked the number of possible arrangements that can be made using the digits of this number.
Now, clearly we can see that we have 7 digits in this number if we are counting the repeated digits differently. We know that n things can be arranged in $n!$ ways if there are n positions. So we have,
$\Rightarrow$ Number of ways in which the digits of the given number can be arranged = $7!$
Now, we can see that the numbers 1, 6 and 4 are repeating twice, so $7!$ would have been the number of possible arrangements in case all the digits would have been different. In cases where we have repetition of letters or digits the effective number of arrangements decreases. The effective number of arrangements will be the number of arrangements obtained in case the digits do not repeat divided by the product of factorials of the number of times each digit is repeating. So we get,
$\Rightarrow$ Effective number of ways in which the digits of the given number can be arranged $=\dfrac{7!}{2!\times 2!\times 2!}=630$.
Hence, there are 630 arrangements of the digits possible.

Note: Note that we cannot always arrange n things in $n!$ ways. An important condition for the arrangement to be $n!$ ways is that we must have n or more than n places where n things have to be arranged. In the above question there are seven digits so they are to be arranged at seven places hence we have obtained $7!$ ways. In case say we have to arrange 7 different things at 5 places then number of arrangements would have been $^{7}{{P}_{5}}=\dfrac{n!}{\left( n-r \right)!}$.