2
$\begingroup$

A word TRAPEZOID is given. We are building words that are 5 letters long, with no repetitions.
a) All possibilites
b) How many words have at least one vowel?
c) How many words have 2 vowels and 3 consonants?

This task is under sector combinations in my book, but shouldn't at least a) be solved with variations?

My solution:
a) variations: $${9*8*7*6*5 = 15120}$$
b) variations (all possibilities - ones without the vowel): $${9*8*7*6*5 - 5*4*3*2*1 = 15000}$$
c) combinations: (2 vowels from 4 and 3 consonants from 5 + choosing the placing of letters): $$\binom{4}{2}*\binom{5}{3}*5!$$
Am I doing something wrong?

  • 1
    a) and b) are OK but not c) since you're not considering where you place the letters. – 2017-01-15
  • 0
    @NeedForHelp I edited the post, is this right? – 2017-01-15
  • 0
    Yes, that update is good. You have "filled your bucket" with five letters, so the $5!$ is needed to order them. – 2017-01-16

1 Answers 1

1

For c) you could proceed like this:

  1. Pick $2$ of the $4$ vowels : $\binom{4}{2}$ choices
  2. Pick $2$ of the $5$ available digits where you place the $2$ vowels chosen in 1.: $\binom{5}{2}$ choices.
  3. Place the vowels at the $2$ digits chosen in 2.: $2!$ choices.
  4. Pick $3$ of the $5$ consonants: $\binom{5}{3}$ choices.
  5. Place the $3$ consonants chosen in 4. at the $3$ remaining digits (those different from the $2$ chosen in 2.): $3!$ choices.

All in all, you have (by the rule of product) $$ \binom{4}{2}\binom{5}{2}\cdot2!\cdot\binom{5}{3}\cdot3!=7200 $$

such words.

  • 0
    Worth noting that this matches the update. – 2017-01-16
  • 0
    @Joffan Yeah, when you have chosen your $2$ vowels and $3$ consonants all you have to do is order them and there is $5!$ ways of doing so. So OP's solution is more efficient than mine! – 2017-01-16