Question

(a) a company that makes crayons is trying to decide which 5 colors to include in a promotional mini - box of 5 crayons. the company can choose the 5 mini - box colors from its collection of 70 colors. how many mini - boxes are possible?
(b) from the 13 albums released by a musician, the recording company wishes to release 9 in a boxed set. how many different boxed sets are possible?

Explanation:

Step1: Use combination formula

The combination formula is \(C(n,r)=\frac{n!}{r!(n - r)!}\), where \(n\) is the total number of items, and \(r\) is the number of items to be chosen.
For part (a), \(n = 70\) and \(r=5\). So \(C(70,5)=\frac{70!}{5!(70 - 5)!}=\frac{70!}{5!×65!}\).
Expand \(70! = 70×69×68×67×66×65!\). Then \(C(70,5)=\frac{70×69×68×67×66×65!}{5!×65!}\).
Since \(5!=5×4×3×2×1 = 120\), \(C(70,5)=\frac{70×69×68×67×66}{120}\).
Calculate \(70×69×68×67×66=(70×66)×(69×67)×68=(4620)×(4623)×68\).
\(4620×4623 = 4620×(4600 + 23)=4620×4600+4620×23=21252000+106260 = 21358260\).
\(21358260×68=21358260×(70 - 2)=21358260×70-21358260×2=1495078200 - 42716520=1452361680\).
\(\frac{1452361680}{120}=12103014\).
For part (b), \(n = 13\) and \(r = 9\). Using the combination formula \(C(13,9)=\frac{13!}{9!(13 - 9)!}=\frac{13!}{9!×4!}\).
Since \(C(n,r)=C(n,n - r)\), \(C(13,9)=C(13,4)\).
\(C(13,4)=\frac{13!}{4!(13 - 4)!}=\frac{13×12×11×10×9!}{4!×9!}\).
\(4!=24\), so \(C(13,4)=\frac{13×12×11×10}{24}\).
\(13×12×11×10 = 17160\), \(\frac{17160}{24}=715\).

Answer:

(a) \(12103014\)
(b) \(715\)