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A certain mathematics contest has a peculiar way of giving prizes. Five people are named as Grand Prize winners, but their finishing order is not listed. Then from among the other entrants, a 6th place, 7th place, 8th place, 9th place, and 10th place winner are each named. If 22 people enter this year, how many complete award announcements are possible?

1 Answer

2 votes

Answer:

19554575040

Step-by-step explanation:

The first five positions to be named as Grand Prize winners are not give any order.

Therefore, the number of ways they can be selected is


_(5)^(22)\textrm{C}


(22!)/(17!5!)

Now the remaining 5 positions are selected from the remaining 17 entries. So they can be ordered as,


_(5)^(17)\textrm{P}


(17!)/(12!)

17 x 16 x 15 x 14 x 13 ways of ordering

Therefore, the number of ways in which the entries can be announced is


(22!)/(17!5!)* 17* 16* 15* 14* 13

= 19554575040 number of ways

Thus the announcement can be made in 19554575040 number of ways.

answered
User Imran Ahmed
by
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