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

(7!)/((2!)(2!)(2!))=5040/(2xx2xx2)=630

#### Explanation:

The number is a seven digit number, so we're dealing with seven digits.

If we were working with the number 1, 234, 567, we'd have 7! ways of ordering the number - the first position could go to any of the seven numbers, then the second position could go to any of the remaining six numbers, the third position could go to any of the five remaining numbers, and so on, giving:

7xx6xx5xx4xx3xx2xx1 = 7! = 5040

But in the number that we have, there are two 1s, two 6s, and two 4s. Each of those groups have internal ordering - for instance, these two numbers are identical:

$9 , 164 , 461 \mathmr{and} 9 , 164 , 461$

even though I've swapped the two 4s around:

$9 , {164}_{\textcolor{\mathmr{and} a n \ge}{1}} , {4}_{\textcolor{b r o w n}{2}} 61 \mathmr{and} 9 , {164}_{\textcolor{b r o w n}{2}} , {4}_{\textcolor{\mathmr{and} a n \ge}{1}} 61$

And so we need to divide by the number of ways we can order these groups of numbers. The number of ways we can order the 1s is 2!, and the same with the 4s and the 6s, and so we get:

(7!)/((2!)(2!)(2!))=5040/(2xx2xx2)=630