## Bowling Pins

After rolling the first ball of a frame in a game of 10-pin bowling, how many different pin configurations can remain (assuming all configurations are physically possible)?
Source: NCTM Mathematics Teacher, September 2006

Solution
A pin configuration is made up of ten pins each of which can be up or down. The number of possible configurations equals $2^{10}=1024$.

Answer: $1024$

Alternative solution
Let $n$ represent the number of pins left standing after a roll. In how many ways can they be left standing?
$\binom{10}{0}=1$ way to leave zero pin (bowl a strike)
$\binom{10}{1}=10$ ways to leave a single pin
$\binom{10}{2}=45$ ways to leave two pins
$\cdots$
$\binom{10}{10}=1$ way to leave all ten pins (gutter ball)
Total number of possible ways
$\binom{10}{0}+\binom{10}{1}+\binom{10}{2}+\binom{10}{3}+\binom{10}{4}+\binom{10}{5}+\binom{10}{6}+\binom{10}{7}+\binom{10}{8}+\binom{10}{9}+\binom{10}{10}=1+10+45+120+210+252+210+120+45+10+1=1024$

Advertisements

## About mvtrinh

Retired high school math teacher.
This entry was posted in Problem solving and tagged , , , , , , , . Bookmark the permalink.