permutations q in half-yearly (1 Viewer)

[5647382910]

This q was in my 3u half yearly but an almost identical q appears in a 4u book so ive posted in this forum:

Question (3 marks) - Eight children, consisting of three boys and five girls are to be seated in a row. In how many ways can this be achieved if no boys are to sit next to each other?
My answer - 5 girls can be arranged in 5! ways, and after this 1st boy can occupy 6 positions, the 2nd 5, the 3rd 4. Hence no of arrangements = 5! x 6 x5 x 4 = 14400.

There is an almost identical q in the 4u fitzpatrick book (q 6a p 226) and the solution is the same as mine. I showed my teacher my solution and he said it was incorrect. I got 0 for this q.

can anyone please verify my solution
thanks in advance

 

[Drongoski]

Yr soln looks right but I'm nervous about such questions. Can u post yr teacher's soln so we can scrutinize it !

 

[study-freak]

This q was in my 3u half yearly but an almost identical q appears in a 4u book so ive posted in this forum:

Question (3 marks) - Eight children, consisting of three boys and five girls are to be seated in a row. In how many ways can this be achieved if no boys are to sit next to each other?
My answer - 5 girls can be arranged in 5! ways, and after this 1st boy can occupy 6 positions, the 2nd 5, the 3rd 4. Hence no of arrangements = 5! x 6 x5 x 4 = 14400.

There is an almost identical q in the 4u fitzpatrick book (q 6a p 226) and the solution is the same as mine. I showed my teacher my solution and he said it was incorrect. I got 0 for this q.

can anyone please verify my solution
thanks in advance

No boys are to sit next to each other and so
total no.of choices=number without restriction-boys sitting next to each other
=8!-6!3!-5!x6x5x3! [6!3! comes from all 3 boys sitting next to one another and 5!x6x5x2! comes from only 2 of the boys sitting next to each other]
=8x7x6!-6x6!-5x6x6!
=(56-6-30)6!
=20x720
=14400 ways
Edit:in the end, got the same answer...
Someone plz correct me if I'm wrong.and edited working lol..

 

[Timothy.Siu]

poster is correct,

5!x6P3

only 2 together would be like 3P2x5!x6P2

 

[study-freak]

Hmm... then I wonder what poster's teacher had got for the answer... lol
It would be terrible if he/she has to end up fixing the whole year's results.

 

[Drongoski]

Not trusting my analytical skills, I've done the problem for case of 3 girls and 2 boys. Based on poster's method there should be 3! x 4x3 = 72 arrangement of the quintet where no 2 boys are together. I made all 5! = 120 possible arrangements and marked off all where no 2 boys are together and got 72 . So it looks like poster is right and teacher is wrong. But poster should also post teacher's 'correct' answer and solution.