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Three out of nine students in the computer club are getting prizes for first, second, and third place in a competition.

How many ways can first, second, and third place be assigned?

A. 84
B. 504
C. 2048
D. 3024


Sagot :

To determine the number of ways to assign first, second, and third place among nine students, we need to calculate the number of permutations of 9 students taken 3 at a time.

A permutation considers the order of selection, which is important in this context as the positions (first, second, third) are distinct.

The formula to calculate permutations [tex]\( _nP_r \)[/tex] is given by:

[tex]\[ _nP_r = \frac{n!}{(n-r)!} \][/tex]

Where:
- [tex]\( n \)[/tex] is the total number of items (in this case, students), which is 9.
- [tex]\( r \)[/tex] is the number of items to choose, which is 3 (for the first, second, and third places).

Applying the values of [tex]\( n \)[/tex] and [tex]\( r \)[/tex]:

[tex]\[ _9P_3 = \frac{9!}{(9-3)!} \][/tex]

Breaking this down:

[tex]\[ _9P_3 = \frac{9!}{6!} \][/tex]

Factorials are calculated as the product of all positive integers up to that number. Thus:

[tex]\[ 9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \][/tex]

[tex]\[ 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 \][/tex]

For convenience, we simplify the calculation by noticing that [tex]\( 9! \)[/tex] can be written as [tex]\( 9 \times 8 \times 7 \times 6! \)[/tex]. Hence,

[tex]\[ _9P_3 = \frac{9 \times 8 \times 7 \times 6!}{6!} = 9 \times 8 \times 7 \][/tex]

Now, multiplying these together:

[tex]\[ 9 \times 8 = 72 \][/tex]

[tex]\[ 72 \times 7 = 504 \][/tex]

Thus, the number of ways to assign the first, second, and third place to the nine students is:

[tex]\[ \boxed{504} \][/tex]