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'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

DOG