Question
how many unique sets of 8 numbers can you make with 64 numbers can you explain it in detail
Answers
GPT-4o mini
To find out how many unique sets of 8 numbers can be formed from a total of 64 numbers, we can use the concept of combinations. Combinations are used here because the order of the numbers does not matter in a set.
We can denote the number of ways to choose \( r \) elements from a set of \( n \) elements as \( C(n, r) \) or \( \binom{n}{r} \). The formula for combinations is given by:
\[
C(n, r) = \frac{n!}{r!(n - r)!}
\]
Where:
- \( n! \) (n factorial) is the product of all positive integers up to \( n \).
- \( r \) is the number of elements to choose.
- \( n \) is the total number of elements to choose from.
In this case:
- \( n = 64 \) (the total numbers available)
- \( r = 8 \) (the number of numbers we want to choose)
Substituting into the formula gives:
\[
C(64, 8) = \frac{64!}{8!(64 - 8)!} = \frac{64!}{8! \cdot 56!}
\]
Now, let's break this down step by step.
1. **Calculate \( 64! \) in terms of cancelled factorials**:
Instead of calculating the full \( 64! \), we can simplify it by canceling \( 56! \):
\[
64! = 64 \times 63 \times 62 \times 61 \times 60 \times 59 \times 58 \times 57 \times 56!
\]
So, we can write:
\[
C(64, 8) = \frac{64 \times 63 \times 62 \times 61 \times 60 \times 59 \times 58 \times 57 \times 56!}{8! \cdot 56!}
\]
The \( 56! \) cancels out:
\[
C(64, 8) = \frac{64 \times 63 \times 62 \times 61 \times 60 \times 59 \times 58 \times 57}{8!}
\]
2. **Calculate \( 8! \)**:
\( 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320 \)
3. **Now perform the multiplication in the numerator**:
\[
64 \times 63 \times 62 \times 61 \times 60 \times 59 \times 58 \times 57
\]
Performing this step-by-step (you can use a calculator for efficiency, but here’s a breakdown):
- \( 64 \times 63 = 4032 \)
- \( 4032 \times 62 = 250104 \)
- \( 250104 \times 61 = 15257744 \)
- \( 15257744 \times 60 = 915464640 \)
- \( 915464640 \times 59 = 53983923840 \)
- \( 53983923840 \times 58 = 3130884878720 \)
- \( 3130884878720 \times 57 = 178755725367040 \)
So, the numerator is \( 178755725367040 \).
4. **Now divide by \( 8! \)**:
\[
C(64, 8) = \frac{178755725367040}{40320}
\]
5. **Perform the final division**:
\[
C(64, 8) = 442112
\]
Thus, the total number of unique sets of 8 numbers that can be formed from 64 numbers is \( \boxed{442211 \text{ (which should be double-checked for computability given the step sizes).}} \)
This represents a vast number of combinations, demonstrating that many selections can be made from a fixed set of elements without regard to the order in which they were selected.
We can denote the number of ways to choose \( r \) elements from a set of \( n \) elements as \( C(n, r) \) or \( \binom{n}{r} \). The formula for combinations is given by:
\[
C(n, r) = \frac{n!}{r!(n - r)!}
\]
Where:
- \( n! \) (n factorial) is the product of all positive integers up to \( n \).
- \( r \) is the number of elements to choose.
- \( n \) is the total number of elements to choose from.
In this case:
- \( n = 64 \) (the total numbers available)
- \( r = 8 \) (the number of numbers we want to choose)
Substituting into the formula gives:
\[
C(64, 8) = \frac{64!}{8!(64 - 8)!} = \frac{64!}{8! \cdot 56!}
\]
Now, let's break this down step by step.
1. **Calculate \( 64! \) in terms of cancelled factorials**:
Instead of calculating the full \( 64! \), we can simplify it by canceling \( 56! \):
\[
64! = 64 \times 63 \times 62 \times 61 \times 60 \times 59 \times 58 \times 57 \times 56!
\]
So, we can write:
\[
C(64, 8) = \frac{64 \times 63 \times 62 \times 61 \times 60 \times 59 \times 58 \times 57 \times 56!}{8! \cdot 56!}
\]
The \( 56! \) cancels out:
\[
C(64, 8) = \frac{64 \times 63 \times 62 \times 61 \times 60 \times 59 \times 58 \times 57}{8!}
\]
2. **Calculate \( 8! \)**:
\( 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320 \)
3. **Now perform the multiplication in the numerator**:
\[
64 \times 63 \times 62 \times 61 \times 60 \times 59 \times 58 \times 57
\]
Performing this step-by-step (you can use a calculator for efficiency, but here’s a breakdown):
- \( 64 \times 63 = 4032 \)
- \( 4032 \times 62 = 250104 \)
- \( 250104 \times 61 = 15257744 \)
- \( 15257744 \times 60 = 915464640 \)
- \( 915464640 \times 59 = 53983923840 \)
- \( 53983923840 \times 58 = 3130884878720 \)
- \( 3130884878720 \times 57 = 178755725367040 \)
So, the numerator is \( 178755725367040 \).
4. **Now divide by \( 8! \)**:
\[
C(64, 8) = \frac{178755725367040}{40320}
\]
5. **Perform the final division**:
\[
C(64, 8) = 442112
\]
Thus, the total number of unique sets of 8 numbers that can be formed from 64 numbers is \( \boxed{442211 \text{ (which should be double-checked for computability given the step sizes).}} \)
This represents a vast number of combinations, demonstrating that many selections can be made from a fixed set of elements without regard to the order in which they were selected.