(a) Five friends are in a netball squad. In each game during the 21-round season, at least 3 of them are picked in the team. Prove that there will be at least 3 matches in which the same three friends are selected to play. (b) How does the answer change if there are six friends instead of 5?
PLS ANSWER FAST

  (a) there are 10 sets of 3 friends, so in 21 games, at least one set must show 3 times

  (b) there are 20 sets of 3 friends, so  in 21 games, at least one set must show 2 times.

Step-by-step explanation:

(a) The number of combinations of 5 things taken 3 at a time is ...

  5C3 = 5!/(3!·2!) = 5·4/2 = 10

There can be 10 games in which the same 3 friends do not show up. There can be 10 more games such that the same 3 friends show up exactly twice. In the 21st game, some set of 3 friends must show up 3 times.

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(b) The number of combinations of 6 things taken 3 at a time is ...

  6C3 = 6!/(3!·3!) = 6·5·4/(3·2) = 20

Hence, there can be 20 games in which the same 3 friends do not show up. In the 21st game, some set of 3 friends will show up a second time.

let's call the solutions x₁ = (5 - 2√7)/3 and x₂ = (5 + 2√7)/3

then the equation is (x-x₁)(x-x₂) = 0

if you fill that in you get a lot of numbers which you can recognize as ax²+bx+c=0.

step-by-step explanation:

you can see that a=1 (there is just an x²)

b = (-5 -2√7 + 2√7 - 5)/3 = -10/3

c = -2√7)/3 ) * (-5+2√7)/3) = -1/3

so your answer is 3x^2 – 10x – 1 = 0