0
$\begingroup$

Question: You have 9 English books, 7 French books and 5 German books. How many ways are there to make a row of three books in which exactly one language is missing (the order of the three books makes a difference) ?

I do not want to use permutations (nPr). My solution is: $$\left(\binom{16}{3} + \binom{14}{3} + \binom{12}{3}\right)\times 3$$ My logic is: pick 3 books out of English & French (16) + pick 3 books out of English & German (14) + pick 3 books out of French & German (12) and multiply by 3 since the order matters (and there are three slots).

Is my answer correct?

Thanks!

  • 0
    What is c here ?2012-04-04
  • 0
    @Thomas: Choice coefficient.2012-04-04
  • 1
    There are $3!=6$ ways to arrange $3$ objects. Apart from that...2012-04-04
  • 1
    No, your answer is not correct. You made no distinction between picking 3 English books or the other situation 2 English and 1 French. The first situation does not work here, the second does. Try picking (1 E and 2 F), (2 E and 1 F), (1 E and 2 G), (2 E and 1 G), (1 F and 2 G), (2 F and 1 G).2012-04-04
  • 0
    Are there any duplicates among the books?2012-04-04

2 Answers 2