To determine the number of different tickets possible in a 5/48 lottery game, where order does not matter, we need to calculate the number of combinations of 48 numbers taken 5 at a time.

The number of combinations of $n$ items taken $k$ at a time is given by the formula: $\binom{n}{k} = \frac{n!}{k!(n-k)!}$

In this case:

• $n = 48$ (total numbers available),
• $k = 5$ (numbers picked for each ticket).

Therefore, the number of different tickets possible is: $\binom{48}{5} = \frac{48!}{5!(48-5)!}$

Let's calculate this value:The number of different tickets possible in the 5/48 lottery game, where order does not matter, is $\binom{48}{5} = 1,712,304$.

Would you like more details on how this calculation was done, or do you have any other questions?

1. What does $\binom{48}{5}$ represent in the context of combinations?
2. How is the factorial operation used in the calculation of combinations?
3. Can you explain why order does not matter in this lottery game?
4. What is the total number of possible outcomes if order did matter in picking 5 numbers from 48?
5. How does the formula for combinations differ from permutations?
6. How many tickets would be possible in a 6/48 lottery game, following the same rules?
7. If someone picks only 1 number in a 48-number lottery game, how many different choices do they have?
8. How does the number of possible tickets change if the number range increased to 60 instead of 48?