Combinatorics and counting

2 free worked examples

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1. Exam PCombinatorics and countingCore

   An actuarial department must send 4 of its 11 analysts to a conference. In how many ways can the group be chosen?

   1. A$44$
   2. B$110$
   3. C$330$
   4. D$1{,}320$
   5. E$7{,}920$

   Solution

   1. The group is unordered - sending analysts A, B, C, D is the same as sending D, C, B, A - so this is a combination.
   2. $\binom{11}{4} = \dfrac{11!}{4!\,7!}$.
   3. $= \dfrac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = \dfrac{7920}{24} = 330$.
   4. The ordered count $P(11,4) = 7920$ is exactly $4! = 24$ times too large, which is why it appears as a distractor.

   Trap. Using the permutation count 7,920, which treats two identical groups in different orders as different selections.

2. Exam PCombinatorics and countingCore

   A password consists of 3 letters (from 26, repeats allowed) followed by 2 digits (from 10, repeats allowed). How many passwords are possible?

   1. A$15{,}600$
   2. B$676{,}000$
   3. C$1{,}404{,}000$
   4. D$1{,}757{,}600$
   5. E$11{,}232{,}000$

   Solution

   1. Each position is chosen independently, so the multiplication rule applies directly.
   2. The letters contribute $26^{3} = 17{,}576$.
   3. The digits contribute $10^{2} = 100$.
   4. $17{,}576 \times 100 = 1{,}757{,}600$ passwords.

   Trap. Using permutations without repetition when the problem explicitly allows repeats.