In how many ways can the digits in the number 9,662,222 be arranged?
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Well, if you insist on arranging the digits in the number 9,662,222, you might end up with quite the repetitive combination. It's like trying to arrange six clowns in a row - doesn't matter how you shuffle them, they'll all still be clowns. So, there's only one way to arrange the digits in 9,662,222 - 9,662,222 itself. In this case, the number is as stubborn as a mule!
To find the number of ways the digits in the number 9,662,222 can be arranged, we need to consider that the digit 2 is repeated five times.
First, let's calculate the total number of arrangements without considering the repeated digit. Since there are 8 digits in total, we have 8 choices for the first position, 7 choices for the second position, 6 choices for the third position, and so on until 1 choice for the last position. Therefore, the total number of arrangements without considering the repeated digit is:
8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40,320
However, since the digit 2 is repeated five times, we need to divide this total by the number of ways we can arrange these repeated digits. Since the five 2's are identical, the number of arrangements for them is given by:
5 x 4 x 3 x 2 x 1 = 120
Therefore, the number of ways the digits in the number 9,662,222 can be arranged is:
40,320 / 120 = 336.
So, there are 336 different ways the digits can be arranged.
To find the number of ways the digits in the number 9,662,222 can be arranged, we need to consider that this number has repeated digits (2 and 6) and a total of 7 digits.
We can use the concept of permutations with repetition to solve this problem. In general, the formula for permutations with repetition can be expressed as:
P(n; n1, n2, ..., nk) = (n!) / (n1! * n2! * ... * nk!)
Where:
- n is the total number of objects to arrange
- n1, n2, ..., nk are the repetitions of each object (n1 times object 1, n2 times object 2, etc.)
- ! denotes the factorial operation
In our case, we have the following numbers of repetitions:
- 6: repeated 3 times
- 2: repeated 4 times
We can plug these values into the formula:
P(7; 3, 4) = (7!) / (3! * 4!)
Calculating this expression:
P(7; 3, 4) = (7 * 6 * 5 * 4 * 3 * 2 * 1) / (3 * 2 * 1 * 4 * 3 * 2 * 1)
P(7; 3, 4) = 35
Therefore, there are 35 different ways the digits in the number 9,662,222 can be arranged.
Normally, with 7 distinct items, there are 7! = 5040 ways to order them.
Since there are 2 6's and 4 2's, their order is indistinguishable. So, there are only
7! / 2!4! = 105 ways
which can be distinguished from each other.