# Permutations Maths Question

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#1
there are 11 candles 3 blue 2 red and 6 green
Find number of different arrangements if:
i) 1 red candle at each end
ii) blue candles together and red candles not together
0
2 months ago
#2
(Original post by papie)
there are 11 candles 3 blue 2 red and 6 green
Find number of different arrangements if:
i) 1 red candle at each end
ii) blue candles together and red candles not together
Any ideas? For i) what happens to the red candles and what's left?
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#3
(Original post by mqb2766)
Any ideas? For i) what happens to the red candles and what's left?
well for i) we are left with 9 candles that can rearrange themselves in 9! ways. because of repetition of blue and green candles, this becomes 9!/ 3! * 6! . right?
0
2 months ago
#4
(Original post by papie)
well for i) we are left with 9 candles that can rearrange themselves in 9! ways. because of repetition of blue and green candles, this becomes 9!/ 3! * 6! . right?
Yes.
0
2 months ago
#5
(Original post by papie)
well for i) we are left with 9 candles that can rearrange themselves in 9! ways. because of repetition of blue and green candles, this becomes 9!/ 3! * 6! . right?
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#6
(Original post by mqb2766)
well that's what i did for b)
treated 3 Blue candles as 1 object so arranging in a line: BBB G G G G G G
for arranging the 6 Green and Blue candles: 7! but 6 identical Green candles so 7!/6! .
the 3 Blue candles can arrange themselves in 3! ways. now it becomes (7! / 6!) * 3!
the 2 Red candles have to be separate so they can take any positions in between the 3 Blue and the Green candles. there are 8 spots so
8P2 / 2! (because 2 identical red candles right?) => this part i'm not sure.
overall i get (8P2) / 2! * 7!/ 6! * 3! = 1176 .
0
2 months ago
#7
(Original post by papie)
well that's what i did for b)
treated 3 Blue candles as 1 object so arranging in a line: BBB G G G G G G
for arranging the 6 Green and Blue candles: 7! but 6 identical Green candles so 7!/6! .
the 3 Blue candles can arrange themselves in 3! ways. now it becomes (7! / 6!) * 3!
the 2 Red candles have to be separate so they can take any positions in between the 3 Blue and the Green candles. there are 8 spots so
8P2 / 2! (because 2 identical red candles right?) => this part i'm not sure.
overall i get (8P2) / 2! * 7!/ 6! * 3! = 1176 .
If treat the 3B as a single block, so have 1B, 2R and 6G. How many possibilities? Think about the B&R placing and use the green as filler.
You don't want two reds together, so subtract the total with 2R together from the previous total.
Last edited by mqb2766; 2 months ago
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#8
(Original post by mqb2766)
If treat the 3B as a single block, so have 1B, 2R and 6G. How many possibilities? Think about the B&R placing and use the green as filler.
You don't want two reds together, so subtract the total with 2R together from the previous total.
9!/ (6! * 2!) * 3! . Is this part correct?
0
2 months ago
#9
(Original post by papie)
9!/ (6! * 2!) * 3! . Is this part correct?
For which part? I get
9*8*7/2
For allocating the single blue block (9 positions) and 2 reds (8*7/2 positions).
0
#10
(Original post by mqb2766)
For which part? I get
9*8*7/2
For allocating the single blue block (9 positions) and 2 reds (8*7/2 positions).
actually i'm confused, i did it like this and i think it's the correct answer:
8P2 / 2! = red candles taking any 8 spots
7!/ (6! * 3!) = 6 green and 1 blue candle (6 G and 3 B repeating)
3! = the 3 Blue candles rearranging among themselves in the group
so multiplying altogether gives me 196 which think is the answer. What do you think ?
0
2 months ago
#11
(Original post by papie)
actually i'm confused, i did it like this and i think it's the correct answer:
8P2 / 2! = red candles taking any 8 spots
7!/ (6! * 3!) = 6 green and 1 blue candle (6 G and 3 B repeating)
3! = the 3 Blue candles rearranging among themselves in the group
so multiplying altogether gives me 196 which think is the answer. What do you think ?
Yes it's 252-56.
There are a few ways of looking at the problem.
1
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