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Sagot :
To determine the number of ways to choose and organize two people out of six, where two specific people are given responsibilities, we can use the concept of permutations.
Permutations are used when the order matters, which is the case here since one person is holding the sign and another is pointing to the sign.
The formula for permutations when choosing and ordering [tex]\( k \)[/tex] out of [tex]\( n \)[/tex] is given by:
[tex]\[ P(n, k) = \frac{n!}{(n - k)!} \][/tex]
In this problem:
- [tex]\( n \)[/tex] is 6 (the total number of people),
- [tex]\( k \)[/tex] is 2 (the number of people chosen to hold responsibilities).
Let's apply these values to the permutation formula:
[tex]\[ P(6, 2) = \frac{6!}{(6 - 2)!} = \frac{6!}{4!} \][/tex]
Now, we can simplify by calculating the factorial expressions:
[tex]\[ 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 \][/tex]
[tex]\[ 4! = 4 \times 3 \times 2 \times 1 \][/tex]
When we divide these two, the [tex]\(4!\)[/tex] terms cancel out:
[tex]\[ \frac{6!}{4!} = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{4 \times 3 \times 2 \times 1} = 6 \times 5 = 30 \][/tex]
So, the number of ways to organize two people out of six with specific responsibilities is:
[tex]\[ 30 \][/tex]
Therefore, the number of ways this can happen is [tex]\( \boxed{30} \)[/tex].
Permutations are used when the order matters, which is the case here since one person is holding the sign and another is pointing to the sign.
The formula for permutations when choosing and ordering [tex]\( k \)[/tex] out of [tex]\( n \)[/tex] is given by:
[tex]\[ P(n, k) = \frac{n!}{(n - k)!} \][/tex]
In this problem:
- [tex]\( n \)[/tex] is 6 (the total number of people),
- [tex]\( k \)[/tex] is 2 (the number of people chosen to hold responsibilities).
Let's apply these values to the permutation formula:
[tex]\[ P(6, 2) = \frac{6!}{(6 - 2)!} = \frac{6!}{4!} \][/tex]
Now, we can simplify by calculating the factorial expressions:
[tex]\[ 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 \][/tex]
[tex]\[ 4! = 4 \times 3 \times 2 \times 1 \][/tex]
When we divide these two, the [tex]\(4!\)[/tex] terms cancel out:
[tex]\[ \frac{6!}{4!} = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{4 \times 3 \times 2 \times 1} = 6 \times 5 = 30 \][/tex]
So, the number of ways to organize two people out of six with specific responsibilities is:
[tex]\[ 30 \][/tex]
Therefore, the number of ways this can happen is [tex]\( \boxed{30} \)[/tex].
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