This task we can solve in several ways:
How many combinations of a 6-digit number are between 000001 to 999999?
How many combinations of a 6-digit number between 000001 and 999999 contain the digit 1 or multiples of digit 1? The 6-digit number is between 000001 to 999,999. The combinations must contain at least 1 digit of “1” or multiples of the digit “1” .
Solution:
To find the number of combinations of 6-digit numbers containing at least one digit of “1” or multiples of digit “1”, we can use complementary counting. This involves calculating the total number of 6-digit numbers and subtracting the number of 6-digit numbers that do not contain “1” or any of its multiples (2, 3, 4, 5, 6, 7, 8, and 9).
The number of 6-digit numbers between 000001 and 999999 is 999,999 – 1 + 1 = 999,999.
Next, we need to find the total number of 6-digit numbers that do not contain “1” or any of its multiples. This means the digits can only be 0 or multiples of 10 (which are just 0). Since there’s only 1 choice for each digit, there will be 1^6 = 1 possible number: 000000. However, this number is not between 000001 and 999999, so there are 0 numbers with this property.
Therefore, the number of 6-digit numbers containing at least one digit of “1” or multiples of digit “1” is:
999,999 – 0 = 999,999.
Combinations
Six numbers can be chosen from a set of 10 numbers, assuming that the numbers can be repeated and order doesn’t matter.
To calculate the number of different combinations of 6 numbers, we can use the formula for combinations, which is:
nCr = n! / (r! * (n-r)!)
where n is the total number of items, r is the number of chosen items, and ! represents the factorial function.
In this case, we want to choose 6 numbers from a set of numbers. Assuming that the numbers can be repeated and order doesn’t matter, we have the following:
If you have a set of numbers from 0 to 9, you have 10 numbers (n = 10). To find the number of combinations of choosing 6 numbers (k = 6) from this set, we can use the same formula I mentioned earlier:
C(n, k) = n! / (k! * (n-k)!)
In this case:
C(10, 6) = 10! / (6! * (10-6)!) C(10, 6) = 10! / (6! * 4!)
Now, we can calculate the factorials:
10! = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720 4! = 4 × 3 × 2 × 1 = 24
Plug these values into the formula:
C(10, 6) = (10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1) / (720 × 24)
Simplify the expression by canceling out common factors:
C(10, 6) = (10 × 9 × 8 × 7) / (4 × 3 × 2 × 1)
Now, calculate the result:
C(10, 6) = 210
So, there are 210 combinations of choosing 6 numbers from a set of 10 numbers (0 to 9).
Therefore, 210 combinations of 6 numbers can be chosen from 10 numbers, assuming that the numbers can be repeated and order doesn’t matter.
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