Question

A pub has 8 different beers on tap. On 4 different days each week they
offer a discount price on one of the beers. The manager is deciding
which beers to discount in a given week, suppose that they only want
to choose the beers, not the specific day each is on special.
(a) If they want to select 4 different beers, how many different choices
are there?
(b) Suppose that they now allow the possibility of selecting the same
beer on multiple days, how many choices are there for selecting
discounts for the week? This means they are choosing some num-
ber of beers, and the number of days each is on special, for a total
of 4. For example one choice might be East End Lager for 2 days
and Adelaide Bitter for 2 days.
(c) If they are still allowing the same beer to be chosen multiple
times, but two of the beers are premium and they want premium
beers to be on special on either 1 or 2 of the 4 days, then how
many choices are there?

Answer 1

Part a

$8C4 = \frac{8!}{ (4! * (8 - 4)!)} \\ \mathrm{Add\:the\:numbers:}\:8-4=4\\ =\frac{8!}{4!\cdot \:4!}\\ \mathrm{Cancel\:the\:factorials}:\quad \frac{n!}{\left(n-m\right)!}=n\cdot \left(n-1\right)\cdots \left(n-m+1\right),\:\quad \:n>\:m\\ \frac{8!}{4!}=8\cdot \:7\cdot \:6\cdot \:5\\ =\frac{8\cdot \:7\cdot \:6\cdot \:5}{4!}\\ =70$

Part b

$\frac{8!}{2!\left(8-2\right)!}\\ \mathrm{Add\:the\:numbers:}\:8-2=6\\ =\frac{8!}{2!\cdot \:6!}\\ \mathrm{Cancel\:the\:factorials}:\quad \frac{n!}{\left(n-m\right)!}=n\cdot \left(n-1\right)\cdots \left(n-m+1\right),\:\quad \:n>\:m\\ \frac{8!}{6!}=8\cdot \:7\\ =28$

Part C

$\frac{8!}{1!\left(8-1\right)!}+\frac{8!}{2!\left(8-2\right)!}+\frac{8!}{4!\left(8-4\right)!}\\ =\frac{192}{24}+\frac{672}{24}+\frac{1680}{24}\\ \mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}\\ =\frac{192+672+1680}{24}\\ \mathrm{Add\:the\:numbers:}\:192+672+1680=2544\\ =\frac{2544}{24}\\ \mathrm{Divide\:the\:numbers:}\:\frac{2544}{24}=106\\ =106$