FMM1 - Revision history

Admin at 21:54, 16 November 2019

2019-11-16T21:54:01Z

← Older revision		Revision as of 21:54, 16 November 2019
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	{{~~Collapse~~\|1=~~Discussion text to~~ be put ~~into box~~.~~\|bg~~=~~#500\|bg2~~=~~#005~~}}		{{FMM
			\|title=Spider Socks and Shoes

			\|problem=
			A spider has one sock and one shoe for each of its eight legs. In how many different orders can the spider put on its socks and shoes, assuming that, on each leg, the sock must be put on before the shoe?

			\|solution=
			Label each of the spider's legs 1 through 8.

			Label the "put on a sock" operation as a, "put on a shoe" operation as b.

			Now, a possible arrangement of ways of putting on socks and shoes is a permutation of the 16 symbols, a1 b1 a2 b2 ... a8 b8 subject to the condition that ak precedes bk for k = 1..8. Our goal is to count the number of these permutations.

			The total number of permutations of the 16 symbols (no conditions) is 16!

			Among those 16! possibilities, exactly half of them have a1 before b1, and the other half have b1 before a1. So, only half of the total permutations fit our criteria for leg number 1. We are down to 16!/2 possibilities.

			But we can apply the same logic for leg 2 - of those 16!/2 possibilities, exactly half of them have a2 before b2, and the other half have b2 before a2. So, only half of those permutations fit our criteria. We are down to 16!/2^2 possibilities.

			We apply the same logic to all 8 legs, to obtain our final answer of

			<math>
			\dfrac{16!}{2^8} = 81,729,648,000
			</math>

			}}

Admin at 21:48, 16 November 2019

2019-11-16T21:48:20Z

← Older revision		Revision as of 21:48, 16 November 2019
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	{{Collapse\|1=Discussion text to be put into box.\|bg=#500\|bg2=#005}}		{{Collapse\|1=Discussion text to be put into box.\|bg=#500\|bg2=#005}}

	~~{{FMM\|title=Spider Socks and Shoes\|problem=A problem\|solution=A solution}}~~

Admin at 21:48, 16 November 2019

2019-11-16T21:48:05Z

← Older revision		Revision as of 21:48, 16 November 2019
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			{{Collapse\|1=Discussion text to be put into box.\|bg=#500\|bg2=#005}}

	{{FMM\|title=Spider Socks and Shoes\|problem=A problem\|solution=A solution}}		{{FMM\|title=Spider Socks and Shoes\|problem=A problem\|solution=A solution}}

	~~{{Collapse\|1=Discussion text to be put into box.\|bg=#500\|bg2=#005}}~~

Admin at 21:47, 16 November 2019

2019-11-16T21:47:52Z

← Older revision		Revision as of 21:47, 16 November 2019
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	{{FMM\|title=Spider Socks and Shoes\|problem=A problem}}		{{FMM\|title=Spider Socks and Shoes\|problem=A problem\|solution=A solution}}

			{{Collapse\|1=Discussion text to be put into box.\|bg=#500\|bg2=#005}}

Admin at 21:45, 16 November 2019

2019-11-16T21:45:01Z

← Older revision		Revision as of 21:45, 16 November 2019
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	{{FMM\|title=Spider Socks and Shoes\|problem=A problem~~\|solution=A solution~~}}		{{FMM\|title=Spider Socks and Shoes\|problem=A problem}}

Admin: Replaced content with "{{FMM|title=Spider Socks and Shoes|problem=A problem|solution=A solution}}"

2019-11-16T21:44:12Z

Replaced content with "{{FMM|title=Spider Socks and Shoes|problem=A problem|solution=A solution}}"

← Older revision		Revision as of 21:44, 16 November 2019
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	{{FMM		{{FMM\|title=Spider Socks and Shoes\|problem=A problem\|solution=A solution}}
	\|title=Spider Socks and Shoes
	\|problem=A spider has one sock and one shoe for each of its eight legs. In how many different orders can the spider put on its socks and shoes, assuming that, on each leg, the sock must be put on before the shoe?

	\|solution=

	~~Label each of the spider's legs 1 through 8.~~

	~~Label the "put on a sock" operation as a, "put on a shoe" operation as b.~~

	Now, a possible arrangement of ways of putting on socks and shoes is a permutation of the 16 symbols, a1 b1 a2 b2 ... a8 b8 subject to the condition that ak precedes bk for k = 1..8. Our goal is to count the number of these permutations.

	~~The total number of permutations of the 16 symbols (no conditions) is 16!~~

	Among those 16! possibilities, exactly half of them have a1 before b1, and the other half have b1 before a1. So, only half of the total permutations fit our criteria for leg number 1. We are down to 16!/2 possibilities.

	But we can apply the same logic for leg 2 - of those 16!/2 possibilities, exactly half of them have a2 before b2, and the other half have b2 before a2. So, only half of those permutations fit our criteria. We are down to 16!/2^2 possibilities.

	~~We apply the same logic to all 8 legs, to obtain our final answer of~~

	~~<math>~~
	~~\dfrac{16!}{2^8} = 81,729,648,000~~
	~~</math>~~

	}}

Admin at 21:43, 16 November 2019

2019-11-16T21:43:48Z

← Older revision		Revision as of 21:43, 16 November 2019
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	{{FMM\|		{{FMM
	\|title=Spider Socks and Shoes		\|title=Spider Socks and Shoes
	\|problem=A spider has one sock and one shoe for each of its eight legs. In how many different orders can the spider put on its socks and shoes, assuming that, on each leg, the sock must be put on before the shoe?		\|problem=A spider has one sock and one shoe for each of its eight legs. In how many different orders can the spider put on its socks and shoes, assuming that, on each leg, the sock must be put on before the shoe?

Admin: Created page with "{{FMM| |title=Spider Socks and Shoes |problem=A spider has one sock and one shoe for each of its eight legs. In how many different orders can the spider put on its socks and s..."

2019-11-16T21:36:35Z

Created page with "{{FMM| |title=Spider Socks and Shoes |problem=A spider has one sock and one shoe for each of its eight legs. In how many different orders can the spider put on its socks and s..."

New page

{{FMM|
|title=Spider Socks and Shoes
|problem=A spider has one sock and one shoe for each of its eight legs. In how many different orders can the spider put on its socks and shoes, assuming that, on each leg, the sock must be put on before the shoe?

|solution=

Label each of the spider's legs 1 through 8.

Label the "put on a sock" operation as a, "put on a shoe" operation as b.

Now, a possible arrangement of ways of putting on socks and shoes is a permutation of the 16 symbols, a1 b1 a2 b2 ... a8 b8 subject to the condition that ak precedes bk for k = 1..8. Our goal is to count the number of these permutations.

The total number of permutations of the 16 symbols (no conditions) is 16!

Among those 16! possibilities, exactly half of them have a1 before b1, and the other half have b1 before a1. So, only half of the total permutations fit our criteria for leg number 1. We are down to 16!/2 possibilities.

But we can apply the same logic for leg 2 - of those 16!/2 possibilities, exactly half of them have a2 before b2, and the other half have b2 before a2. So, only half of those permutations fit our criteria. We are down to 16!/2^2 possibilities.

We apply the same logic to all 8 legs, to obtain our final answer of

<math>
\dfrac{16!}{2^8} = 81,729,648,000
</math>

}}