Statistics Question: 10 flowers. 5 red, 3 yellow, 2 purple.. # different arrangements?

Question by lovingkindness: Statistics Question: 10 flowers. 5 red, 3 yellow, 2 purple.. # different arrangements?
I’m not sure how to solve this problem. If someone could explain the steps I’d be very grateful!

“If you were planting 10 flowers in a row and 5 of them were red, 3 of them were yellow, and 2 of them were purple, in how many different looking arrangements could the 10 flowers be arranged?”

Best answer:

Answer by wiseguy
Dear fournier,

There are ten positions in the row you are planting. Imagine that you decide where to plant the red flowers first, and that flowers are not distinguished by any other features . Then there are C(10, 5) different placements possible for these five flowers, where C(n,k) = n! / [k! (n - k)!] (known as the binomial coefficient, which is the number of ways that exactly k items can be chosen from a set of n items). This leaves five positions for the other flowers. If you decide where to plant the yellow flowers next, then there are C(10 – 5, 3) = C(5, 3) different placements for these flowers, given the positions of the red flowers. The remaining purple flowers go into the last two positions, so there is only one placement left for them, given the positions of the red and yellow flowers. (You can express this mathematically as C(10 – 5 – 3, 2) = C(2, 2) = 1.) If you multiply together these ways of placing each color of flower, then you obtain the total number of different-looking arrangements:

C(10, 5) C(10 – 5, 3) C(10 – 5 – 3, 2)
= C(10, 5) C(5, 3) C(2, 2)
= (10! / [5! (10 - 5)!]) (5! / [3! (5 - 3)!]) (2! / [2! (2 - 2)!])
= (10! / [5! 5!]) (5! / [3! 2!]) (2! / [2! 0!])
= 10! / (5! 3! 2!)
= (10)(9)(8)(7)(6) / [(3)(2)(2)]
= (10)(9)(4)(7)
= 2520.

Notice that there is no requirement that you decide to start by placing the red flowers. For instance, if you begin with the yellow flowers, then the purple, and finally the red, the total number of different-looking arrangements is the same:

C(10, 3) C(10 – 3, 2) C(10 – 3 – 2, 5)
= C(10, 3) C(7, 2) C(5, 5)
= (10! / [3! 7!]) (7! / [2! 5!]) (5! / [5! 0!])
= 10! / (3! 2! 5!)
= 2520.

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