To solve the problem, we can approach it by first calculating the total number of ways to form a committee of 3 partners from the 41 partners (16 senior + 25 junior), and then subtract the cases where no junior partners are selected.

Step 1: Total number of ways to form a committee

The total number of ways to choose 3 partners from 41 is given by the combination formula:

$\binom{41}{3} = \frac{41 \times 40 \times 39}{3 \times 2 \times 1} = 10660$

Step 2: Number of ways where no junior partners are chosen

In this case, all 3 partners must be senior partners. The number of ways to choose 3 partners from the 16 senior partners is:

$\binom{16}{3} = \frac{16 \times 15 \times 14}{3 \times 2 \times 1} = 560$

Step 3: Number of ways with at least one junior partner

To find the number of ways where at least one junior partner is chosen, we subtract the cases with no junior partners from the total number of ways to form the committee:

$\text{Ways with at least one junior partner} = \binom{41}{3} - \binom{16}{3} = 10660 - 560 = 10100$

Thus, the number of ways to form a committee with at least one junior partner is 10,100.

Would you like further details or have any questions?

Related Questions:

1. How many ways can exactly two junior partners be selected for the committee?
2. What is the probability that all selected committee members are junior partners?
3. How many committees can be formed if the firm adds more senior or junior partners?
4. How would the result change if the committee size was increased to 4 partners?
5. If one senior partner must be on the committee, how many ways can the committee be formed?

Tip:

Remember that combinations are useful when the order of selection does not matter, while permutations are for when order does matter.