David, Tommy, Jack, and Michael

First, let us calculate the total number of ways any three people can be selected for the positions.

$\dbinom{9}{3} = 84$

Next, we calculate the number of cases where none of the three (Tommy, Jack and Michael) are selected.

$\dbinom{9 - 3}{3} = \dbinom{6}{3} = 20$

So, the number of ways to select people such that at least one of the three is selected is given by:

$84 - 20 = \boxed{64}$

First, compute for the all the possible number ways of selecting 3 persons to be on the Head of the Council out of the original 9, that is:

$\dbinom{9}{3} = 84$

Then, compute for the number ways of selecting the 3 persons excluding Tommy, Jack & Michael. Now, that's:

$\dbinom{9-3}{3} = 20$

Find the difference of the two, and you'll get:

$84 - 20 = \boxed{64}$

This number shows the number of possible ways of including at least one of the 3 to the Head Council.

First, we derive that he could choose the council in $\frac{9 \times 8 \times 7} {3 \times 2}=84$ . Next, we derive the number of ways that he could choose the council without Tommy, Jack nor Michael. This is $\frac{6 \times 5 \times 4}{3 \times 2}=20$ Thus the amount of ways that he could choose the council with at least one of the three is $84-20=\boxed{64}$ .

First we need to find how many ways can Tommy be chosen only. Since there are 6 other unknown people, we need to know what $5 + 4 + 3 + 2 + 1$ equals which is $15$ . Multiply 15 by 3, since the question is asking about 3 people. We get an answer of $45$ . Now we need to know how many different combinations can have 2 of the known people. There are 6 unknown and 3 different ways to arrange the 3 known people in pairs. $6 * 3$ equals $18$ . Now, we need to know how many different ways can all of them be arranged. This is, obviously, $1$ . Now we need to add $45$ , $18$ and $1$ and you get an answer of $64$ .

9 people, subsets of 3 people:

$9 \choose 3$ $= 84$ combinations

subtract the total number of combinations that don't have Tommy, Jacky, or Michael (there are 6 other people and, again, we are dealing with subsets of 3 people)

$84 -$ $6 \choose 3$ $= \boxed{64}$

Number of ways in which David can choose at 3 people out of 9, such that at least one of Tommy, Jack and Michael is chosen = Total number of ways to choose 3 people out of 9 - number of ways to choose 3 people out of 9, such that none of Tommy, Jack and Michael are chosen.

Total number of ways to choose 3 people out of 9 = $_{3}^{9}\textrm{C}$

$= \frac{9!}{3!} \times 6!$

$= 84$

Number of ways to choose 3 people out of 9, such that none of Tommy, Jack and Michael are chosen = Number of ways to choose 6 people out of 9 = $_{3}^{6}\textrm{C}$

$\frac{6!}{3! \times 3!}$

$= 20$

Hence, answer = $84 - 20 = \boxed{64}$

There are $\binom{9}{3}=84$ total ways to choose 3 people from 9 applicants without restrictions. Now, we find the number of ways to choose 3 people such that none of Tommy, Jack, or Michael are chosen. We are choosing 3 people from 6 people which can be done in $\binom{6}{3}=20$ ways. Therefore there are $84-20=\boxed{64}$ ways of choosing a group of 3 people such that the restriction is satisfied.

This is an alternate way (a bit more complicated): We consider 2 groups. One group is Tommy, Jack, and Micheal; another group is the rest (6 people).

-One out of three on the head:

$\dbinom{3}{1}$ * $\dbinom{6}{2}$ = 45 ways

-Two out of three on the head:

$\dbinom{3}{2}$ * $\dbinom{6}{1}$ = 18 ways

-Three out of three on the head: 1 way

Using rule of sum, the final answer will be 45 + 18 + 1 = 64 ways

$sum _{ i=1 }^{ 3 }{ (3Ci)*(6C(3-i)) } =64$

$\dbinom{9}{3} - \dbinom{6}{3} = \boxed {64}$

(9!÷6!÷3!)-(6!÷3!÷3!)=64 ans

selecting one of 3= 3c1=3 * selecting 2 from remaining 6= 6C2=15 +3C2*6C1+1

Tommy, Jack and Micheal are all chosen, that is 1 way. Tommy and Jack and 1 of the 6 (9-3) remaining people, that is 6 ways. Tommy and Micheal and 1 of the 6 remaining people, that is 6 ways. Jack and Micheal and 1 of the 6 remaining people, that is 6 ways. Tommy or Jack or Micheal with 2 of the 6 remaining people, that is 3* C2 in 6 = 3* 6!/(4!2!) = 45 Add everything...45+3*6+1=45+19=64!

9C3 - 6C3 = 64

I did not know the rules so i had to right evry single propability hhhhhhhhhhhhh :D

9c3-6c3= 84-20=64

You can use the complement of the event $A$ -> $|A|' = |U| - |A|$ Where $A$ is an event that at least one of them is chosen. So $A'$ is none of them is chosen. And $U$ is the set of total ways to select 3 from 9 people.

• Select 3 from 9 people, order doesn't matter, $|U| = C(9,3)$
• Select 3 from 6 people, none of them are selected, $|A'| = C(6,3)$

$|A|' = |U| - |A| => |A| = C(9,3) - C(6,3) = \boxed{64}$

We can approach the problem with 2 ways. 1) Find the number of possible ways we could have at least 1 of them (1, 2, or all 3) 2) The complement approach: Total number of ways - Total ways without any of them.

The second one will be the easier task. There will be the remaining 6 to choose from the applicants for the complement set. Because of that, there are 6C3 ways to have these. 6C3 = 6 * 5 * 4 / 3 * 2 * 1 = 20

The total number of ways is 9C3 = 9 * 8 * 7 / 3 * 2 * 1= 84

The final answer will be 84 - 20 = 64 ways to have at least one of them to be part of the committee's head council