Group A consists of 7 boys and 3 girls, while group B consists of 6 boys and 5 girls. The number of ways, 4 boys and 4 girls can be invited for a picnic if 5 of them must be from group $A$ and the remaining 3 from group $B$, is equal to:

Step 1: Define variables and conditions

First, let's clearly define the composition of each group and the requirements for the picnic. We need to select a total of 8 people (4 boys and 4 girls), with 5 coming from Group A and 3 from Group B.

Step 2: Determine possible distributions of boys and girls from Group A

Since 5 members must be selected from Group A, we need to find combinations of boys ($b_A$) and girls ($g_A$) such that $b_A + g_A = 5$. Also, the number of boys cannot exceed 4 (total boys needed) and the number of girls cannot exceed 4 (total girls needed). Considering the available members in Group A (7 boys, 3 girls), the possible distributions are: 2 boys and 3 girls, 3 boys and 2 girls, or 4 boys and 1 girl.

Step 3: Determine corresponding distributions for Group B

For each case from Group A, we determine the required number of boys ($b_B$) and girls ($g_B$) from Group B. The total number of boys must be 4, and the total number of girls must be 4. Also, the sum $b_B + g_B$ must equal 3, as 3 members are selected from Group B.

Step 4: Calculate combinations for each case

We calculate the number of ways for each case using combinations. For each case, we multiply the number of ways to choose boys from Group A, girls from Group A, boys from Group B, and girls from Group B. Remember that $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.

Step 5: Sum the combinations for all cases

Finally, we sum the number of ways from all possible cases to get the total number of ways to invite 4 boys and 4 girls for the picnic under the given conditions.

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