Permutation question in discrete mathematics. At least 1 out 3 members (P) from a total of 10 members

Question

Im doing a question out of Discrete and Combinatorial mathematics by Grimaldi (4th Edition). I'm stuck on one of the questions and am trying to find an alternative way of doing it, that is not in the solution. As background I'm a Computer Science second year student but I just enjoy maths so I'm going further than I may need to for my degree.

The Board of Directors of a pharmaceutical corporation has 10 members. An upcoming stockholders meeting is scheduled to approve a new slate of company officers (chosen from the 10 board members). How many different slates consisting of a President, Vice President, Secretary and Treasurer can the board present to the stockholders for their approval?

Thats simply 10!/6!= 5040

Three members of the board of directors are Physicians. How many slates have at least 1 physician appearing?

The solution in the textbook is very smart and simple.

There are 7 x 6 x 5 x 4= 840 slates without Physicians, so 5040-840=420 
slates with at least 1 physicians

Im asking for an alternative way to solve this. I know that there are 2520 slates of exactly 1 physician.

4 x [3 x 7 x 6 x 5]= 2520

I multiply by 4 here because there are 4 positions available, so that 1 physician can be in any of those 4 positions

If I were to find it for exactly 2 physicians it would be:

? x [3 x 2 x 7 x 6]

However I do not know how to proceed further.

Answer 1

As you say, there are $3\cdot4\cdot7\cdot6\cdot5$ slates with exactly one physician.

To get the number of slates with exactly 2 physicians, we can select the 2 physicians, assign them to their offices, and then fill the other 2 offices; and this can be done in $\dbinom{3}{2}\cdot4\cdot3\cdot7\cdot6$ ways.

To get the number of slates with all 3 physicians, we can assign the physicians to their offices and then fill the remaining office; and this can be done in $4\cdot3\cdot2\cdot7$ ways.

This gives a total of $4\cdot3\cdot7\cdot[30+18+2]=4200$ possibilities.

Answer 2

I think the question is rather how we got this $4$ in our $4\times[3\times 7\times 6\times 5]$ answer.

There are $3$ physicians for one position or $P(3,1)$, and $P(7,3)$ ways to fulfill the rest of the positions. And since there are $4$ positions total arrangements like $3\times 7\times 5\times 6$, $7\times 3\times 5\times 6$, $7\times 6\times 3\times 5$, $7\times 6\times 5\times 3$ represent different slates, so we have to use the rule of sum to find the number of ways to accomplish task. $3\times 7\times 5\times 6 + 7\times 3\times 5\times 6 + 7\times 6\times 3\times 5 + 7\times 6\times 5\times 3 = 4[3\times 7\times 5\times 6]$

Solving by the rule of product. There are $4$ possible outcomes for choosing position (president, vice, etc, etc), $3$ for choosing physician, and $7\times 6\times 5$ for leftovers. We have three procedures $P(4,1), P(3,1), P(7,3)$, or $4, 3, 7\times 6\times 5$. We have outcomes, one to all (representing different arrangements). $P(4,1)\times P(3,1)\times P(7,3)$

For exactly $2$, and $3$ there would be overcounting, which needs to be reducing either in positions, or in physicians, see first post.