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 :

We need to form a group of 8 people (4 boys, 4 girls) such that 5 are from Group A (7B, 3G) and 3 are from Group B (6B, 5G).

First, we will distribute boys and girls from Group A.

Let Boys from A = x and Girls from A = y

Then, x + y = 5

Now, from Group B:

Boys = 4 - x

Girls = 4 - y

Now, the possible cases are:

Case 1: x = 4, y = 1

From A: $$^7C_4\times^3C_1=35\times3=105$$

From B: $$^6C_0\times^5C_3=1\times10=10$$

Total = 105*10 = 1050

Case 2: x = 3, y = 2

From A: $$^7C_3\times^3C_2=35\times3=105$$

From B: $$^6C_1\times^5C_2=6\times10=60$$

Total = 105*60 = 6300

Case 3: x = 2, y = 3

From A: $$^7C_2\times^3C_3=21\times1=21$$

From B: $$^6C_2\times^5C_1=15\times5=75$$

Total = 21*75 = 1575

Sum = 1050 + 6300 + 1575 = 8925