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The-number-of-ways-in-which-8-different-flowers-can-be-strung-to-form-a-garland-so-that-4-particular-flowers-are-never-separated-is-

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Question Number 11889 by murtaza8 last updated on 04/Apr/17
The number of ways in which 8 different flowers can be strung to form a garland so that 4 particular flowers are never separated is
$$\mathrm{The}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ways}\:\mathrm{in}\:\mathrm{which}\:\mathrm{8}\: \\ $$$$\mathrm{different}\:\mathrm{flowers}\:\mathrm{can}\:\mathrm{be}\:\mathrm{strung}\:\mathrm{to} \\ $$$$\mathrm{form}\:\mathrm{a}\:\mathrm{garland}\:\mathrm{so}\:\mathrm{that}\:\mathrm{4}\:\mathrm{particular} \\ $$$$\mathrm{flowers}\:\mathrm{are}\:\mathrm{never}\:\mathrm{separated}\:\mathrm{is} \\ $$
Answered by ajfour last updated on 04/Apr/17
(((5−1)!)/2)(4!)
$$\frac{\left(\mathrm{5}−\mathrm{1}\right)!}{\mathrm{2}}\left(\mathrm{4}!\right) \\ $$
Commented by ajfour last updated on 04/Apr/17
Commented by mrW1 last updated on 04/Apr/17
Can you please explain your solution? I think he meant that 4 special flowers should always chosen together.
$${Can}\:{you}\:{please}\:{explain}\:{your}\:{solution}? \\ $$$${I}\:{think}\:{he}\:{meant}\:{that}\:\mathrm{4}\:{special}\:{flowers} \\ $$$${should}\:{always}\:{chosen}\:{together}. \\ $$
Commented by mrW1 last updated on 04/Apr/17
if we treat these 4 flowers as one super flower. 5 flowers form the garland, there are 5! way. To replace this super flower with our 4 always−together−flowers, there are 4 ways. So the solution is 5!4!=2880
$${if}\:{we}\:{treat}\:{these}\:\mathrm{4}\:{flowers}\:{as}\:{one}\:{super}\:{flower}. \\ $$$$\mathrm{5}\:{flowers}\:{form}\:{the}\:{garland},\:{there}\:{are} \\ $$$$\mathrm{5}!\:{way}.\:{To}\:{replace}\:{this}\:{super}\:{flower} \\ $$$${with}\:{our}\:\mathrm{4}\:{always}−{together}−{flowers}, \\ $$$${there}\:{are}\:\mathrm{4}\:{ways}.\:{So}\:{the}\:{solution}\:{is} \\ $$$$\mathrm{5}!\mathrm{4}!=\mathrm{2880} \\ $$
Commented by mrW1 last updated on 04/Apr/17
Sorry, I understand that the garland should consist of 4 flowers.
$${Sorry},\:{I}\:{understand}\:{that}\:{the}\:{garland} \\ $$$${should}\:{consist}\:{of}\:\mathrm{4}\:{flowers}. \\ $$
Commented by ajfour last updated on 04/Apr/17
but those four can be arranged among themselves.. for 4+(1group)=5 flowers have a cicular permutation of (((5−1)!)/2) . And the four not to be separated should,i believe, have a permutation of 4! .
$${but}\:{those}\:{four}\:{can}\:{be}\:{arranged} \\ $$$${among}\:\:{themselves}.. \\ $$$${for}\:\mathrm{4}+\left(\mathrm{1}{group}\right)=\mathrm{5}\:{flowers} \\ $$$${have}\:{a}\:{cicular}\:{permutation}\:{of} \\ $$$$\frac{\left(\mathrm{5}−\mathrm{1}\right)!}{\mathrm{2}}\:.\:{And}\:{the}\:{four}\:{not}\:{to}\:{be} \\ $$$${separated}\:{should},{i}\:{believe},\:{have}\:{a}\: \\ $$$${permutation}\:{of}\:\:\mathrm{4}!\:. \\ $$
Commented by ajfour last updated on 04/Apr/17
if there are 3 different flowers to form a garland we have (((3−1)!)/2) =1 way and that′s true.
$${if}\:{there}\:{are}\:\mathrm{3}\:{different}\:{flowers}\: \\ $$$${to}\:{form}\:{a}\:{garland}\:{we}\:{have} \\ $$$$\frac{\left(\mathrm{3}−\mathrm{1}\right)!}{\mathrm{2}}\:=\mathrm{1}\:{way}\:{and}\:{that}'{s}\:{true}. \\ $$
Commented by prakash jain last updated on 04/Apr/17
If we need to arrange n object in a circular permutation than number of arragements are (n−1)! like seating on a round table. ABC, CAB, BCA are same but ACB is different from ABC. For arrangements that can turned upside down like a garland so clockwise and anticlockwise are equivalent, number of permutation are (((n−1)!)/2)
$$\mathrm{If}\:\mathrm{we}\:\mathrm{need}\:\mathrm{to}\:\mathrm{arrange}\:{n}\:\mathrm{object}\:\mathrm{in}\:\mathrm{a} \\ $$$$\mathrm{circular}\:\mathrm{permutation}\:\mathrm{than}\:\mathrm{number} \\ $$$$\mathrm{of}\:\mathrm{arragements}\:\mathrm{are}\:\left({n}−\mathrm{1}\right)!\:\mathrm{like} \\ $$$$\mathrm{seating}\:\mathrm{on}\:\mathrm{a}\:\mathrm{round}\:\mathrm{table}. \\ $$$$\mathrm{ABC},\:\mathrm{CAB},\:\mathrm{BCA}\:\mathrm{are}\:\mathrm{same}\:\mathrm{but} \\ $$$$\mathrm{ACB}\:\mathrm{is}\:\mathrm{different}\:\mathrm{from}\:\mathrm{ABC}. \\ $$$$\mathrm{For}\:\mathrm{arrangements}\:\mathrm{that}\:\mathrm{can}\:\mathrm{turned} \\ $$$$\mathrm{upside}\:\mathrm{down}\:\mathrm{like}\:\mathrm{a}\:\mathrm{garland}\: \\ $$$$\mathrm{so}\:\mathrm{clockwise}\:\mathrm{and}\:\mathrm{anticlockwise} \\ $$$$\mathrm{are}\:\mathrm{equivalent}, \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{permutation}\:\mathrm{are}\:\frac{\left(\mathrm{n}−\mathrm{1}\right)!}{\mathrm{2}} \\ $$
Commented by mrW1 last updated on 06/Apr/17
thanks!
$${thanks}! \\ $$

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