Helpppppppppp?

A florist has to make a floral arrangement. She has 6 banksias, 5 wattles and 4 waratahs. All the flowers of each kind are different. In how many ways can the florist make a bunch of 10 flowers if she has to use at least 3 of each kind?

Helpppppppppp?
Call the flowers 6A, 5B, 4C. All the flowers of each kind are different =%26gt; each combination of them is unique.





So Pascal%26#039;s coefficient gives the formula for choosing m objects from a collection of n unique objects:


C(n,m) = n!/(n-m)!m!





Number of ways to make a bunch of 10 flowers if she has to use at least 3 of each kind?


# = Number of ways choosing 4 of A, 3 of B and 3 of C


+ Number of ways choosing 3 of A, 4 of B and 3 of C


+ Number of ways choosing 3 of A, 3 of B and 4 of C





= [C(6,4) * C(5,3) * C(4,3)]


+ [C(6,3) * C(5,4) * C(4,3)]


+ [C(6,3) * C(5,3) * C(4,4)]





= [6!/4!2! * 5!/3!2! * 4!/3!1!]


+ [6!/3!3! * 5!/4!1! * 4!/3!1!]


+ [6!/3!3! * 5!/3!2! * 4!/4!0!]





= [6!/4!2! * 5!/3!2! * 4!/3!1!]


+ [6!/3!3! * 5!/4!1! * 4!/3!1!]


+ [6!/3!3! * 5!/3!2! * 4!/4!0!]





Factorizing out a common factor (6!5!4! / 4!3!3!)





# = (6!5!4! / 4!3!3!) *


( [1/2! * 1/2! * 1/1!]


+ [1/3! * 1/1! * 1/1!]


+ [1/3! * 1/2! * 1/0!] )





= (6!5!4! / 4!3!3!) *


( 1/2 * 1/2 * 1/1]


+ [1/6 * 1/1 * 1/1]


+ [1/6 * 1/2 * 1/1] )





= (6!5!4! / 4!3!3!) * ( 1/4 + 1/6 + 1/12 )





= (6!5!4! / 4!3!3!) * ((3+2+1)/12)


= (6!5!4! / 4!3!3!) * 1/2


= 6!5!4! / 4!3!3! 2


= 6!5! / 3!3! 2


= (6*5*4)(5*4)/2


= (6*5*4)(10)


= 1200



super nanny