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find the number of ways 3 parties can be assigned to 3 of the 12 empty booths of a restaurant
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@mathstudent55
Let A, B, C be the 3 parties and D be considered an empty party or table so one combo is ABCDDDDDDDDD and another combo is ACDDDDDBDDDD etc etc
soo the answer is ?? @jim_thompson5910
there are 12 P 12 = 12! = 479001600 ways to arrange the 12 positions....BUT...D repeats itself (there are 9 copies of D) so you have to divide by 9! to get 479001600/(9!) = 479001600/362880 = 1320
so there are 1320 different ways to assign parties A, B, C to the 12 tables
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