To solve the problem of grouping 10 students into 5 groups of 2, we use the concept of combinations and division by repeated groupings.

Step-by-Step Solution:

1. Total Number of Ways to Pair Up: First, calculate the number of ways to choose 2 students out of 10 for the first group. This is done using combinations: $\binom{10}{2} = \frac{10 \times 9}{2 \times 1} = 45$

2. For the Next Groups: After selecting the first pair, 8 students remain. Now, we choose 2 out of these 8 for the second group: $\binom{8}{2} = \frac{8 \times 7}{2 \times 1} = 28$ Repeating this process for the remaining students: $\binom{6}{2} = 15, \quad \binom{4}{2} = 6, \quad \binom{2}{2} = 1$

3. Multiply All the Combinations: To find the total number of ways to form all the groups: $45 \times 28 \times 15 \times 6 \times 1 = 113400$

4. Account for Group Indistinguishability: Since the order of the groups doesn’t matter, we must divide by the number of ways to arrange the 5 groups. There are $5!$ ways to arrange 5 groups, and $5! = 120$. Thus, the number of ways to group the students is: $\frac{113400}{120} = 945$

Final Answer:

There are 945 ways to group 10 students into 5 groups of 2.

Would you like more details on this approach or have any questions?

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Here are 5 related questions:

1. How would the calculation change if the groups were distinguishable?
2. What if we wanted to form unequal-sized groups?
3. How does the number of groups affect the outcome in similar problems?
4. Can you apply this logic to partition 12 students into 4 groups of 3?
5. How does this problem relate to partition theory in combinatorics?

Tip: Always remember to adjust for overcounting when dealing with indistinguishable groups.