shahrat wrote:helloooooo everybody
can any one plz help me in getting the lgic and answer to this question
before starting a game , a teacher must divide her 10 students into teams of three students ( where order does not matter ) and then assign the remaining student the role of the referee . how many different assignments can she make for the 10 students?
i will appreciate the help
Hi!
Let's attack this question in 3 parts: choosing team 1, choosing team 2 and choosing team 3.
Since order doesn't matter for our choices, we start with combinations:
Team 1: 10 people available, choosing 3, so:
10C3 = 10!/3!7! = 10*9*8/3*2*1 = 720/6 = 120
Team 2: 7 people left, choosing 3, so:
7C3 = 7!/3!4! = 7*6*5/3*2*1 = 7*5 = 35
Teach 3: 4 people left, choosing 3, so:
4C3 = 4!/3!1! = 4/1 = 4
Ref: 1 person left, choosing 1, so only 1 choice!
Whenever we make multiple selections, we multiply the possibilities, giving us:
120 * 35 * 4 * 1 = some big number.
However, that big number isn't the final answer to the question (that's why I haven't done the math yet). We need to recognize that we've counted some groups multiple times.
Here's what I mean:
let's call our 10 students ABCDEFGHIJ.
By our counting method, we could have:
Team 1: ABC
Team 2: DEF
Team 3: GHI
ref: J.
We could also have:
Team 1: DEF
Team 2: GHI
Team 3: ABC
Ref: J.
Now, these are identical assignments, but we've counted them twice. So, we need to factor out our multiples.
Since there are 3 teams, there are 3!=6 different ways to arrange them (123, 132, 213, 231, 312, 321). Therefore, we've counted every assignment 6 times.
So, the final answer to the question is:
(120 * 35 * 4 * 1)/6
= 20 * 35 * 4
= 700 * 4
= 2800
That's a lot of teams! I hope the teacher has a lot of spare time.
