The problem here is calculating the odds of winning a 6-of-6 lottery by using the combination formula for selecting 6 numbers from a total of 53.

The general combination formula is:

$\binom{n}{r} = \frac{n!}{r!(n - r)!}$

In this case:

• $n = 53$: total numbers available,
• $r = 6$: numbers to be chosen.

Step-by-Step Calculation:

1. Set up the formula: $\binom{53}{6} = \frac{53!}{6!(53 - 6)!} = \frac{53!}{6! \cdot 47!}$

2. Simplify the factorial expression: Only the first 6 terms of 53! are needed since the rest (from 47!) cancel out. This leaves: $\frac{53 \times 52 \times 51 \times 50 \times 49 \times 48}{6 \times 5 \times 4 \times 3 \times 2 \times 1}$

3. Perform the multiplication in the numerator and the denominator:

   • Numerator: $53 \times 52 \times 51 \times 50 \times 49 \times 48 = 229,784,800$
   • Denominator: $6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$

4. Divide the results: $\frac{229,784,800}{720} = 22,957,480$

Odds of Winning

The odds of winning are $1 : 22,957,480$.

Answer to Question

The correct choice is:

• 1 : 22,957,480

Would you like further details on the calculation, or do you have any additional questions?

Here are some related questions for further practice:

1. How would the odds change if the total numbers were 60 instead of 53?
2. What are the odds if a lottery requires choosing 5 numbers out of 50?
3. How do combinations differ from permutations in probability?
4. How is probability affected when drawing numbers without replacement?
5. Why does the combination formula use factorials?

Tip: Remember that in combinations, the order of selected items does not matter. This is different from permutations where the order is important.