Reyjidn
Answered

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in a certain general assembly,three major prizes are at stake.In how many ways can be the first,second,and third prizes be drawn from a box containing 120 names?(using the permutation)

Sagot :

Permutation of n taken r at a time expressed as P(n,r) has the formula

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

P(120, 3) = [tex] \frac{120!}{(120-3)!} [/tex] = [tex] \frac{120!}{117!} [/tex] 

By factorial definition, 

120! = 120x119x118x117x116x...x3x2x1

117! =                         117x116x...x3x2x1

So, 

[tex] \frac{120!}{117!} [/tex] = [tex] \frac{120x119x118x117!}{117!} [/tex] (Note that [tex] \frac{117!}{117!} = 1[/tex])

[tex]= \frac{120x119x118x117!}{117!} [/tex]

=[tex]=120x119x118[/tex]

=1685040

Therefore, the first, second, and third prizes can be drawn in 1,685,040 ways.