Okay, so I need to figure out how many ways we can form a subcommittee of 5 members from a Senate committee that has 8 Republicans and 6 Democrats. The key here is that the subcommittee must have at least one member from each party. Let me start by understanding the problem step by step.

First, the total number of members in the committee is 8 Republicans + 6 Democrats = 14 members. We need to choose 5 members from these 14. However, the subcommittee must include at least one Republican and at least one Democrat. That means we can't have all 5 members being Republicans or all 5 being Democrats. So, maybe I should calculate the total number of ways to choose 5 members and then subtract the cases where all are Republicans or all are Democrats. That seems like a common approach for "at least one" problems.

Let me recall the formula for combinations. The number of ways to choose k items from a set of n is given by the combination formula C(n, k) = n! / (k! * (n - k)!). So, the total number of possible subcommittees without any restrictions is C(14, 5). Then, subtract the number of subcommittees that are all Republicans and all Democrats.

Calculating C(14, 5): Let me compute that. 14 choose 5 is 14! / (5! * 9!) = (14*13*12*11*10)/(5*4*3*2*1). Let me calculate that step by step. 14*13=182, 182*12=2184, 2184*11=24024, 24024*10=240240. Then divide by 120 (which is 5!). 240240 / 120 = 2002. So, C(14,5) is 2002.

Now, the number of all-Republican subcommittees. There are 8 Republicans, so we need to choose 5 from them. That's C(8,5). Similarly, the number of all-Democrat subcommittees is C(6,5). Let me compute those.

C(8,5): 8! / (5! * 3!) = (8*7*6)/(3*2*1) = 56. Because 8*7=56, 56*6=336, divided by 6 is 56.

C(6,5): 6! / (5! * 1!) = 6. That's straightforward.

So, the total number of invalid subcommittees (all Republicans or all Democrats) is 56 + 6 = 62.

Therefore, the number of valid subcommittees is total subcommittees minus invalid ones: 2002 - 62 = 1940.

Wait, let me check that again. 2002 total, subtract 56 and 6. Yes, 56 + 6 is 62. 2002 - 62 is indeed 1940. Hmm.

But let me think again. Is there another way to approach this problem? Maybe by considering the different valid distributions of Republicans and Democrats in the subcommittee. Since the subcommittee has 5 members, and at least one from each party, the possible distributions are:

- 1 Republican and 4 Democrats
- 2 Republicans and 3 Democrats
- 3 Republicans and 2 Democrats
- 4 Republicans and 1 Democrat

These are the four cases we need to consider. Then, we can compute the number of ways for each case and sum them up. Let me try this approach to verify.

First, case 1: 1 Republican and 4 Democrats. The number of ways is C(8,1) * C(6,4). C(8,1) is 8. C(6,4) is 15. So, 8*15=120.

Case 2: 2 Republicans and 3 Democrats. C(8,2) * C(6,3). C(8,2)=28. C(6,3)=20. 28*20=560.

Case 3: 3 Republicans and 2 Democrats. C(8,3)*C(6,2). C(8,3)=56. C(6,2)=15. 56*15=840.

Case 4: 4 Republicans and 1 Democrat. C(8,4)*C(6,1). C(8,4)=70. C(6,1)=6. 70*6=420.

Adding all these up: 120 + 560 + 840 + 420. Let's compute:

120 + 560 = 680

680 + 840 = 1520

1520 + 420 = 1940.

Same result as before. So, that's reassuring. So, both methods give 1940. That seems correct.

But wait, let me check once more. Maybe I made a mistake in the combination calculations.

First approach:

Total: C(14,5)=2002.

Invalid: C(8,5)+C(6,5)=56+6=62.

2002-62=1940. Correct.

Second approach:

Sum of all valid cases:

1 Republican: C(8,1)=8, 4 Democrats: C(6,4)=15. 8*15=120.

2 Rep: C(8,2)=28, 3 Dem: C(6,3)=20. 28*20=560.

3 Rep: C(8,3)=56, 2 Dem: C(6,2)=15. 56*15=840.

4 Rep: C(8,4)=70, 1 Dem: C(6,1)=6. 70*6=420.

Total: 120+560=680; 680+840=1520; 1520+420=1940. Correct.

So, both methods confirm that the answer is 1940. Therefore, the number of ways is 1940.

But let me think again if there's any alternative way someone might approach this problem and perhaps make a mistake. For example, someone might forget to subtract the all-Democrat or all-Republican cases, leading to an overcount. But in this case, since we're using the inclusion-exclusion principle, subtracting those cases should be correct. Also, when we considered the different distributions, we covered all possible valid cases, so that's thorough.

Another possible mistake is miscalculating combinations. Let me verify each combination:

C(8,5): 8*7*6*5*4/(5*4*3*2*1) = (8*7*6)/6 = 8*7=56. Correct.

C(6,5): 6. Correct.

C(8,1)=8. Correct.

C(6,4)=15. Correct.

C(8,2)=28. Correct.

C(6,3)=20. Correct.

C(8,3)=56. Correct.

C(6,2)=15. Correct.

C(8,4)=70. Correct.

C(6,1)=6. Correct.

All the individual combination calculations seem correct. So, multiplying and adding them up as above also seems correct.

Alternatively, if someone uses a different method, like generating functions, they might model the problem as (R + D)^5, where R is the number of Republicans and D is the number of Democrats, but since the committee has 8 R and 6 D, but we need to choose 5. Wait, maybe that's complicating it. Alternatively, the generating function for Republicans would be 1 + C(8,1)x + C(8,2)x^2 + ... + C(8,8)x^8, and similarly for Democrats. The coefficient of x^5 in the product of these two generating functions would give the number of ways. But that's essentially the same as considering the different cases. Let me check.

The generating function for Republicans is (1 + x)^8, but since we can only choose up to 5, the relevant terms are up to x^5. Similarly, for Democrats, (1 + x)^6, up to x^5. Then, the product would have terms up to x^10, but we need the coefficient of x^5 in the product. However, since we need at least one from each party, the coefficient of x^5 in the product minus the coefficients where all are R or all are D. Wait, but actually, the generating function approach would automatically subtract those cases because the generating function includes all possible combinations. Wait, no. The coefficient of x^5 in the product of (1 + C(8,1)x + C(8,2)x^2 + ... + C(8,5)x^5) and (1 + C(6,1)x + C(6,2)x^2 + ... + C(6,5)x^5) would give the total number of ways, including all R and all D. But since we need at least one from each party, we need to subtract the cases where all are R (which would be C(8,5)x^5) and all are D (C(6,5)x^5). Then, the coefficient of x^5 would be total - C(8,5) - C(6,5). Which is exactly the same as the first approach. So, that's another way to see it.

Therefore, all methods converge to 1940. I think that's solid. So, the answer should be 1940.
Therefore, the number of ways to form a subcommittee of 5 members with at least one member from each party is $\boxed{1940}$.
</think>