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There was a question in a mock test of an olympiad I gave, which says,
"In how many ways 5 humans and 4 monkeys be seated around a table, such that no two monkeys sit together"
Now, in this question how can I deal with the statement that no two monkeys can sit together.

Mayank M.
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4 Answers4

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Hint first arrange the humans around the table alternatively. This can be done in $5!$ ways . Now automatically no two monkeys are together so they can be arranged in $4!$ ways hence total ways are $5!.4!$

Archis Welankar
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  • I think that you need to take into account symmetric cases around the table. – barak manos Aug 09 '16 at 17:40
  • Can you give an example ? – Archis Welankar Aug 09 '16 at 17:46
  • For example: $H_1M_1H_2M_2H_3M_3H_4M_4H_5$ and $M_4H_5H_1M_1H_2M_2H_3M_3H_4$. – barak manos Aug 09 '16 at 17:49
  • For each given arrangement of the five humans, the four monkeys can be inserted in $5!$ ways. It's the five humans that can be arranged in $4!$ ways (modulo circular symmetry). – Barry Cipra Aug 09 '16 at 18:06
  • the humans can be arranged in $5!\cdot 9$ ways because you have $4!$ orderings (counting cyclic equivalents only once, $5$ choices as to which two are sitting next to each other, and 9 choices for where at the table that human pair starts. – Mark Fischler Aug 09 '16 at 20:00
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Take the monkeys to be distinguishable, the table seats to be distinguishable, and the humans to be distinguishable.

No two monkeys can sit together implies that every monkey has a human on his right (if this was a linear table rather than circular, an end case would have to be considered separately). So we can glue a human to each monkey, thiws can be done in $5\cdot4\cdot3\cdot2=120$ ways. Lable these glued pairs A, B, C, D

Now we can consider pair A as "first" and order pairs B, C, and D in $6$ ways. Having done that, we can decide to slot the leftover human in 4 ways (before A is the same as after D).
Finally, we can start pair A in $9$ possible monkey seats.

$$ 120 \cdot 6 \cdot 4 \cdot 9 = 25920 $$ arrangements.

Mark Fischler
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  • I think I agree with true blue anil's answer. Since this is a circle, all arrangements are modulo circular symmetry, hence your answer should be divided by $9$ (9 animals) which gives $2880$. This is also same numerically as Archis' answer. – Shailesh Aug 22 '16 at 16:13
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I tried to visualize the question and I find the only combination of no two monkeys can sit together is as this(https://i.imgsafe.org/a16562328c.jpg)

Because if you want to switch the seats of monkeys, like this(https://i.imgsafe.org/a170d53318.jpg)
It would still be the exact same permutation as the first one( you can visualize it by rotating the picture).

So only one permutation is allowed, and assume that we have different people and different monkeys, the number of ways would be $5!*4!$

Nick
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Unless otherwise specified, seats at a round table are treated as unnumbered,
hence the formula $(n-1)!$

Taking each living being to be distinct, seat the "chief" somewhere among five chairs,
and arrange the other humans in $4!$ ways.

Four chairs for the monkeys can be inserted in the $5$ gaps between humans $\binom54$ ways, and the monkeys now seated in $4!$ ways

Putting the pieces together, $4!\binom544! = 2880$ ways

NOTE:

Although your header mentions "patterns", there is no such word in the question.

true blue anil
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  • Well I think when question says "in how many ways" it meant the same as "pattern" and we know that the header should be short and informative so I changed a bit of the vocabulary. – Mayank M. Aug 10 '16 at 14:34
  • Patterns and number of ways aren't necessarily the same thing. – true blue anil Aug 10 '16 at 14:47
  • Well I'm curious! How? I mean they grammatically mean nearly the same. – Mayank M. Aug 10 '16 at 15:06
  • Well, just look at this answer on MSE for the difference between a pattern and an arrangement. http://math.stackexchange.com/questions/749141/number-of-ways-of-arranging-7-coloured-blocks-in-patterns – true blue anil Aug 10 '16 at 16:11