This is equal to entry <math>k</math> in row <math>n</math> of Pascal's triangle. Rather than performing the multiplicative calculation, one can simply look up the appropriate entry in the triangle (constructed by additions). For example, suppose 3 workers need to be hired from among 7 candidates; then the number of possible hiring choices is 7 choose 3, the entry 3 in row 7 of the above table (taking into consideration the first row is the 0th row), which is <math> \tbinom{7}{3}=35 </math>.<ref>{{Cite web |url=http://5010.mathed.usu.edu/Fall2018/HWheeler/probability.html |access-date=2023-06-01 |website=5010.mathed.usu.edu|title=Pascal's Triangle in Probability}}</ref>
== Relation to binomial distribution and convolutions ==
When divided by <math> 2^n</math>, the <math> n</math>th row of Pascal's triangle becomes the [[binomial distribution]] in the symmetric case where <math> p = \tfrac{1}{2}</math>. By the [[central limit theorem]], this distribution approaches the [[normal distribution]] as <math> n</math> increases. This can also be seen by applying [[Stirling's formula]] to the factorials involved in the formula for combinations.
This is related to the operation of [[discrete convolution]] in two ways. First, polynomial multiplication corresponds exactly to discrete convolution, so that repeatedly convolving the sequence <math> \{ \ldots, 0, 0, 1, 1, 0, 0, \ldots \}</math> with itself corresponds to taking powers of <math> x + 1</math>, and hence to generating the rows of the triangle. Second, repeatedly convolving the distribution function for a [[random variable]] with itself corresponds to calculating the distribution function for a sum of ''n'' independent copies of that variable; this is exactly the situation to which the central limit theorem applies, and hence results in the normal distribution in the limit. (The operation of repeatedly taking a convolution of something with itself is called the [[convolution power]].)
