Combinatorics (1 Viewer)

[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.

 

[justletmespeak123]

Sure! Let's break it down step by step.

Step 1: Total Number of Positions

Since the boys and girls must alternate, the arrangement will look like this:

• Boy, Girl, Boy, Girl, Boy, Girl, Boy, Girl (or vice versa, starting with a girl).

This gives us two possible patterns:

1. Boy, Girl, Boy, Girl, Boy, Girl, Boy, Girl
2. Girl, Boy, Girl, Boy, Girl, Boy, Girl, Boy

But, since we are not yet concerned with which pattern (starting with a boy or a girl), let's just focus on the alternating pattern for now.

Step 2: Number of Ways to Arrange Boys and Girls

There are 4 boys and 4 girls, and they need to be arranged in these alternating positions.

• Boys can be placed in 4 specific spots (since the alternating pattern requires the boys to be in certain places). The number of ways to arrange the boys in these spots is ( 4! ) (4 factorial).
• Girls can be placed in the remaining 4 spots. The number of ways to arrange the girls is also ( 4! ) (4 factorial).

So, without any further restrictions, the total number of ways to arrange the boys and girls would be:
[
4! \times 4!
]

Step 3: Account for John and Sally

Now, we need to incorporate the condition that John is next to Sally.

• Treat John and Sally as a single "block" or "unit" because they must be next to each other. This means we now have 3 boys and 3 girls to arrange (excluding John and Sally).
• John and Sally can be arranged within their "block" in 2 ways: John can be on the left or Sally can be on the left.

Now, we need to arrange the remaining 3 boys and 3 girls in the remaining positions. Since we still need to alternate, we can follow the same logic as before:

• There are 3 boys and 3 girls to be arranged in alternating positions. The number of ways to do this is ( 3! \times 3! ).

Step 4: Total Number of Arrangements

Now, we can calculate the total number of arrangements by multiplying the possibilities together:

1. Number of ways to arrange the 3 remaining boys and 3 remaining girls: ( 3! \times 3! ).
2. Number of ways to arrange John and Sally within their "block": 2 ways (John can be left or right).
3. Total number of ways to arrange the blocks of boys and girls, including John and Sally: ( 3! \times 3! \times 2 ).

Thus, the total number of arrangements is:
[
(3! \times 3! \times 2) \times 4! \times 4!
]


Now, let's calculate this:
[
3! = 6, \quad 4! = 24
]
So,
[
(6 \times 6 \times 2) \times (24 \times 24) = 72 \times 576 = 41472
]

Final Answer:

The total number of arrangements is 41,472.