Combinatorics (3 Viewers)

[yiaghertop]

Four girls and Four boys are to stand in a row. Find the number of arrangements if Boys and Girls alternate and John is next to Sally. Any ideas?

Answer: 504

 

[f7eeting]

hi! the way id do this is id consider arranging alternating boys and girls first which would look like:
4!x4!x2 (where multiplying for 2 accounts for the separate cases where either a boy could be first or a girl)
then, since we're told that a specific boy and a specific girl MUST sit next to each other, id remove them from the sample space of boys and girls, leaving us with:
3!x3!x2
but since there are numerous ways we can arrange john and sally together, you have to account for that as well, so i multiply 3!x3! by 7 (this can be seen easily if you draw out 8 boxes and see how many ways you can shuffle 2 objects across)
and overall i get:
3!x3!x7x2
and to break it down:
3!x3! (arranges the other boys and girls that are not john or sally) x7 (arranging the possible positions john and sally could be in) x 2 (counts for the case where a boy is first or a girl)

and while i was doing this, i was thinking about counting the cases where john and sally could be switched but then realised you cant do that because then it wouldnt alternate between boys and girls. a bit trivial but just putting it out there incase it helps clarify anything
hope this helps!

 

[yiaghertop]

hi! the way id do this is id consider arranging alternating boys and girls first which would look like:
4!x4!x2 (where multiplying for 2 accounts for the separate cases where either a boy could be first or a girl)
then, since we're told that a specific boy and a specific girl MUST sit next to each other, id remove them from the sample space of boys and girls, leaving us with:
3!x3!x2
but since there are numerous ways we can arrange john and sally together, you have to account for that as well, so i multiply 3!x3! by 7 (this can be seen easily if you draw out 8 boxes and see how many ways you can shuffle 2 objects across)
and overall i get:
3!x3!x7x2
and to break it down:
3!x3! (arranges the other boys and girls that are not john or sally) x7 (arranging the possible positions john and sally could be in) x 2 (counts for the case where a boy is first or a girl)

and while i was doing this, i was thinking about counting the cases where john and sally could be switched but then realised you cant do that because then it wouldnt alternate between boys and girls. a bit trivial but just putting it out there incase it helps clarify anything
hope this helps!

Thank you so much, I really appreciate your help!!

 

[Luca26]

Start with the case where the alternating pattern is B-G-B-G-B-G-B-G:

• Place John in position 1 (1 way)
  • in which case Sarah has to be position 2 (1 way)
  • other three boys can be positioned in 3! ways
  • and same for other three girls
• So, John in position 1 gives 1 x 1 x 3! x 3! = (3!)² arrangements
• Now, John could be in any of the other three B positions (3 ways)
  • in which case Sarah has to be in one of two positions adjacent to John (2 ways)
  • other three boys can be positioned in 3! ways
  • and same for other three girls
• So, John not in position 1 gives 3 x 2 x 3! x 3! = (3!)³ arrangements
• So, this case yields (3!)² + (3!)³ = (3!)²(1 + 3!) = 7 x 6² = 252 arrangements

The alternative case with arrangements G-B-G-B-G-B-G-B will, by placing Sarah analogously to John above, yield 252 arrangements

Thus, there are a total of 504 arrangements.