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# The number of ways in which 8 different flowers can be

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CEO
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The number of ways in which 8 different flowers can be [#permalink]

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03 Dec 2007, 09:43
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The number of ways in which 8 different flowers can be seated to form a garland so that 4 particular flowers are never separated is:

A. 4!4!
B. 288
C. 8!/4!
D. 5!4!
E. 8!4!
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03 Dec 2007, 09:55
D.

Ways in which you can arrange the 4 different flowers = 4!
Ways in which you can arrage the set of 4 flowers +the rest of the flowers = 5!

Although I always get confused with this questions. If it doesn't say that the 4 flowers must be in the same order, can I assume that I can flip them around? Don't know why I always get confused by this.
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21 Dec 2007, 23:25
bmwhype2 wrote:
The number of ways in which 8 different flowers can be seated to form a garland so that 4 particular flowers are never separated is:
A) 4!4!
B) 288
C) 8!/4!
D) 5!4!
E) 8!4!

4 flowers which are always together can be considered as one SET,

Therefore we have to arrange one SET ( 4 flowers ) and 4 other flowers into a garland.

Which means, 5 things to be arranged in a garland.

(5-1)!

And the SET of flowers can arrange themselves within each other in 4! ways.

Therefore

(5-1)!*(4!)

But, Garland, looked from front or behind does not matter. Therefore the clockwise and anti clockwise observation does not make difference.

Therefore

(5-1)! * (4!)/ 2

= 288.

What is the OA?
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22 Dec 2007, 00:07
its a circular perm

OA is A
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22 Dec 2007, 02:02
bmwhype2 wrote:
its a circular perm

OA is A

That means clock-wise and anti-clock wise combinations differ from each other. Hmmmm.....perhaps I should visualize more! I just imagined the flowers look the same, whether looked from front of behind in a garland. But their colors would differ!
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05 Jan 2008, 09:00
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Expert's post
A

1. We have 5 different things: the group of 4 flowers and 4 separate flowers. 5P5=5!
2. to arrange the group of 4 flowers we have 4P4=4! ways. So, 5!*4!
3. circular symmetry means that variants with "circular shift" are the same variant. We can make 5 "circular shifts". Therefore, N=5!*4!/5=4!*4!
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25 Aug 2008, 13:51
1
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bmwhype2 wrote:
The number of ways in which 8 different flowers can be seated to form a garland so that 4 particular flowers are never separated is:
A) 4!4!
B) 288
C) 8!/4!
D) 5!4!
E) 8!4!

[1234]5678

Assume that 1234 are alwasy together So. we can arrange themselves in 4! ways.
X5678
Now treat [1234]=X one single group we have 5 flower snad arrange in circular way= (5-1)!

4!*4!
A.
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27 Sep 2009, 21:38
The number of ways in which 8 different flowers can be seated to form a garland so that 4 particular flowers are never separated is:
A) 4!4!
B) 288
C) 8!/4!
D) 5!4!
E) 8!4!

Soln. I too go with A.
(5-1)! * 4!
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28 Oct 2009, 10:56
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I feel 4! * 4! is not the final answer to this question. This number should be divided by 2 because a single garland when turned around gives us a different arrangement, but its still the same garland.

Answer: 4! * 4!/2 = 288
B

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27 Jan 2010, 01:00
if 4 flowers must be toghter, we can think that at first we must seat that flowers in 5 seats, in that case ther are 5! cases, but we have 4flowers which in every case of 5! we can arrange its in 4! case, so there are 5!*4! cases
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27 Jan 2010, 04:38
Expert's post
2
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samrus98 wrote:
I feel 4! * 4! is not the final answer to this question. This number should be divided by 2 because a single garland when turned around gives us a different arrangement, but its still the same garland.

Answer: 4! * 4!/2 = 288
B

This is a good point.

There are two cases of circular-permutations:

1. If clockwise and anti clock-wise orders are different, then total number of circular-permutations is given by $$(n-1)!$$.

2. If clock-wise and anti-clock-wise orders are taken as not different, then total number of circular-permutations is given by $$\frac{(n-1)!}{2!}$$.

Specific garland (as I understand) when turned around has different arrangement, but its still the same garland as Samrus pointed out. So clock-wise and anti-clock-wise orders are taken as not different.

Hence we'll have the case 2: $$\frac{(5-1)!*4!}{2}=288$$
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Re: combinatorics - garland   [#permalink] 27 Jan 2010, 04:38
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