1
$\begingroup$

the question is as follows: suppose a dj has 5 english hits and 4 french hits.

  • in how many ways can these songs be played if no two english hits should follow each other?
  • in how many ways can these songs be played if no two french hits should follow each other?

For the first question, the order should be EFEFEFEFE. The E's can be permuted and so can the F's. The total is therefore $5!4! = 2880$.

For the second question, the previous question is included. But so is the possibility that 2 english hits and 3 english hits follow each other.

  • Only one English hit each time: 2880 possibilities.
  • two English hits back to back: take this pair as 'one' song. Than we need to divide 4 French songs and 4 English songs. Hence either the french songs are in the odd spots or in the even spots. Moreover, there are 4 places where we can put the extended English song and afterwards, we can permute all English songs as well as the French songs. This gives a total of $2 \cdot 4! \cdot 4 \cdot 5! = 8 \cdot 2880$ possibilities.
  • two extended songs of 2 english songs. This gives 7 spots to fill in. The french songs should be in the odd spots, the english songs in the even spots. There are 2 out of three spots to place an extended mix. Moreover, the english songs and french songs can be permuted, so we obtain $3 \cdot 2880$ possibilities.
  • three english songs back to back. The french songs are now in the odd spots, the English songs (2 english, one extended) in the even spots. There are 3 places to put the extended song, and we can permute all english and all french songs. This gives a total of $3 \cdot 2280$ possibilities to arange these songs.

This gives a total of $(1 + 8 + 3 + 3)\cdot 2880 = 43200$ possibilities, which is the answer in the back of my book.

question: However, this reasoning is not concise at all and I needed to make many case-distinctions (I also forgot the 2 pairs of 2 songs-case at first). Is there a more elegant way to solve this question?

$\endgroup$
2
$\begingroup$

First question is correct.

Second question: there are $5!$ ways to place $5$ English songs. Now there are $6$ places to place $4$ French songs, which is the permutation $P(6,4)=\frac{6!}{2!}$. Hence, the required answer is: $$5!\cdot \frac{6!}{2!}=43200.$$

$\endgroup$
4
  • 2
    $\begingroup$ Somebody who is not already familiar with this method might not understand that the six places are the four spaces between successive English songs and the ends of the row: $\square E \square E \square E \square E \square E \square$. $\endgroup$ Apr 15 '19 at 9:43
  • 1
    $\begingroup$ @N.F.Taussig, right, but one should realize that since the 5 English songs are already placed, then 4 French songs can be placed in-between or outside them. $\endgroup$
    – farruhota
    Apr 15 '19 at 10:02
  • $\begingroup$ @farruhota thanks a lot, together with N. F. Taussig's comment this makes perfectly sense. $\endgroup$
    – Student
    Apr 15 '19 at 10:05
  • $\begingroup$ He knows it backwards and forwards. I have also learned much from him. Good luck. $\endgroup$
    – farruhota
    Apr 15 '19 at 10:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.