Another counting question

Toast · 2006-12-22 06:58:13

Four boys and three girls are to be seated in a row. In how many ways can this be done if there is a boy at one end and a girl at the other end?
I did:
[4][5][4][3][2][1][3] (each box representing number of branches in the tree. 4 is on the left since 1 of 4 boys must sit on the left and 3 is on the right because 1 of 3 girls must sit on the right)
= 4 × 5 × 4 × 3 × 2 × 1 × 3 = 1440 combinations.

However, shouldn't 1440 be multiplied by 2 since it didn't specifically say 'boy at the left end and girl at the right end', so you could really just flip it around (girl at the left and boy at the right) and get another 1440 combinations = 2880 combinations?
The answers said 1440, but I'm a bit doubtful.

Last edited by Toast (2006-12-22 07:08:45)

mathsyperson · 2006-12-22 07:11:55

I agree with your answer, and that the 1440 should have been multiplied by 2 because the positions of the boy and girl at the ends weren't specified.
I think it's just that the book said one thing and meant another.

fgarb · 2006-12-22 07:20:44

Sounds like it's just an ambiguous question. When it said "at one end", it probably just meant "at this one end", when you could just as easily interpret it as meaning "an end (either one)". Just goes to show teachers need to be careful in how they phrase their questions.

pi man · 2006-12-22 11:55:02

I agree with the 2880. I used a different approach. There are 7! = 5040 different ways to arrange the 7 kids. Now subtract out the number of ways that there are either 2 boys or 2 girls on the ends.

2 Boys in the end seats: 4C2 * 2! * 5! = 6 * 2 * 120 = 1440
4C2 denotes (4 choose 2) and is the number of ways to pick the 2 boys for the end seats. Multiply that by 2! because that's the number of ways to arrange those 2 boys. The 5! is the number of ways to arrange the 5 kids (2 boys, 3 girls) in the inner seats.

2 Girls in the end seats: 3C2 * 2! * 5! = 3 * 2 * 120 = 720
3C2 denotes (3 choose 2) and is the number of ways to pick the 2 girls for the end seats. Multiply that by 2! because that's the number of ways to arrange those 2 girls. The 5! is the number of ways to arrange the 5 kids (4 boys, 1 girl) in the inner seats.

5040 - 1440 - 720 = 2880

Is there a standard way to denote (X choose Y) in text? I'll have to read through Latex notes to see if it's capable of doing the standard notation.

Stanley_Marsh · 2006-12-22 16:22:34

the first seat can have one of the 4 boys , the last one can have one of the 3 girls , so there are 4x3=12 ways , after chosing one boy and one girl ,there are 5 people left , so it has 5!=120 combination. In total there are 120x12=1440 arrangment

Stanley_Marsh · 2006-12-22 16:24:43

"there is a boy at one end and a girl at the other end" . I assume they dont switch end ~lol

krassi_holmz · 2006-12-24 06:56:28

I aggree with tha answer 2880, and my approach is different.
Let's exclude the two ends for a moment. 1 boy and 1 girl are standing there, so for the other 5 seats we have 5 persons, which makes 5! ways for sitting.
Now in the one end can seat 4 boys and in the other - 3 girls, but because we have 2 ends, we'll multiply by 2, so in total:
2.3.4.5!=2880.

Math Is Fun Forum

#1 2006-12-22 06:58:13

Another counting question

#2 2006-12-22 07:11:55

Re: Another counting question

#3 2006-12-22 07:20:44

Re: Another counting question

#4 2006-12-22 11:55:02

Re: Another counting question

#5 2006-12-22 16:22:34

Re: Another counting question

#6 2006-12-22 16:24:43

Re: Another counting question

#7 2006-12-24 06:56:28

Re: Another counting question

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