Answer:
Step-by-step explanation:
if we have n people and we want to do groups of 5, the total number of different combinations is:
[tex]c = \frac{n!}{(n - 5)!5!}[/tex]
find the smallest n such c > 365.
let's do it by brute force:
if n = 5, we have:
[tex]c = 5C_5 = 5[/tex]
if n = 6
[tex]c =6C_5 =6[/tex]
if n = 7
[tex]c = 7C_ 5= 21[/tex]
if n = 8
[tex]c = 8C_5 = 56[/tex]
if n = 9
[tex]C = 9!/(4!*5!) = 126[/tex]
so 9 is not enough, let's see N = 10
C = 10!/(5!*5!) = 252
10 is not enough, let's see with 11.
C = 11!/(6!*5!) = 462
So you need at least 11 members in the club.
Then the minimum number of members such we have more than 365 combinations is 11 members
Client need al least 11 members in the club.
According to the question,
Group of five members, the combination will be:
→ [tex]c = \frac{n!}{(n-5)! \ 5!}[/tex]
By using Brute force,
If n = 5,
c = 5,
C₅ = 5
If n = 6,
c = 6,
C₅ = 6
If n = 7,
c = 7,
C₅ = 21
If n = 8,
c = 8,
C₅ = 56
Now,
If n = 9,
→ C = [tex]\frac{9!}{(4!\times 5!)}[/tex]
= 126
and, If n = 10,
→ C = [tex]\frac{10!}{(5!\times 5!)}[/tex]
= 252
and, If n = 11,
→ C = [tex]\frac{11!}{(6!\times 5!)}[/tex]
= 462
Then perhaps the minimal number of individuals required to have more than 365 possible combinations seems to be 11.
Thus the response above is correct.
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