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Sagot :
To determine the number of ways to play three songs out of a total of eight in sequence, we need to calculate the number of permutations of 8 items taken 3 at a time.
The formula for permutations [tex]\( P(n, r) \)[/tex] is given by:
[tex]\[ P(n, r) = \frac{n!}{(n-r)!} \][/tex]
Here, [tex]\( n = 8 \)[/tex] (the total number of songs) and [tex]\( r = 3 \)[/tex] (the number of songs to be played in sequence).
Let's insert the values into the formula:
[tex]\[ P(8, 3) = \frac{8!}{(8-3)!} \][/tex]
This simplifies to:
[tex]\[ P(8, 3) = \frac{8!}{5!} \][/tex]
Since [tex]\( 8! \)[/tex] (8 factorial) is the product of all positive integers up to 8, and [tex]\( 5! \)[/tex] (5 factorial) is the product of all positive integers up to 5, we can write it as:
[tex]\[ 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \][/tex]
[tex]\[ 5! = 5 \times 4 \times 3 \times 2 \times 1 \][/tex]
Now, [tex]\( 5! \)[/tex] (the product of the numbers from 5 down to 1) will cancel out from both the numerator and the denominator, leaving us with:
[tex]\[ P(8, 3) = \frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{5 \times 4 \times 3 \times 2 \times 1} = 8 \times 7 \times 6 \][/tex]
So we calculate:
[tex]\[ 8 \times 7 = 56 \][/tex]
[tex]\[ 56 \times 6 = 336 \][/tex]
Therefore, the number of ways to play three of the eight songs in sequence is:
[tex]\[ \boxed{336} \][/tex]
The formula for permutations [tex]\( P(n, r) \)[/tex] is given by:
[tex]\[ P(n, r) = \frac{n!}{(n-r)!} \][/tex]
Here, [tex]\( n = 8 \)[/tex] (the total number of songs) and [tex]\( r = 3 \)[/tex] (the number of songs to be played in sequence).
Let's insert the values into the formula:
[tex]\[ P(8, 3) = \frac{8!}{(8-3)!} \][/tex]
This simplifies to:
[tex]\[ P(8, 3) = \frac{8!}{5!} \][/tex]
Since [tex]\( 8! \)[/tex] (8 factorial) is the product of all positive integers up to 8, and [tex]\( 5! \)[/tex] (5 factorial) is the product of all positive integers up to 5, we can write it as:
[tex]\[ 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \][/tex]
[tex]\[ 5! = 5 \times 4 \times 3 \times 2 \times 1 \][/tex]
Now, [tex]\( 5! \)[/tex] (the product of the numbers from 5 down to 1) will cancel out from both the numerator and the denominator, leaving us with:
[tex]\[ P(8, 3) = \frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{5 \times 4 \times 3 \times 2 \times 1} = 8 \times 7 \times 6 \][/tex]
So we calculate:
[tex]\[ 8 \times 7 = 56 \][/tex]
[tex]\[ 56 \times 6 = 336 \][/tex]
Therefore, the number of ways to play three of the eight songs in sequence is:
[tex]\[ \boxed{336} \][/tex]
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