Sweets and Chocolates!

Probability Level 4

At a party, each person brings 1 packet of sweets and 1 chocolate bar. The packets of sweets and the chocolate bars have to be shared out among the people at the party such that each person gets one (whole) packet of sweets and one (whole) chocolate bar, but no one gets the sweets or chocolate that they brought with them. If there are 7 people at the party, how many ways are there to do this?

Details and Assumptions:

The sweets and chocolate bar that each person brings is different from the sweets and chocolate bar that anyone else brings.

Image Credit: Wikimedia Candies

The answer is 3437316.

1 solution

Brian Charlesworth
Mar 5, 2015

This is a "double" derangement problem, involving two separate and independent derangements of $7$ items each. As the number of derangements of a set of $7$ items is $1854$ , the desired solution is $1854^{2} = \boxed{3437316}.$

So , there 's a long shot . If there are" n " derangement , involving n separate and independent derangements of m objects each then the desired number of ways is the n the power of the derangements of n objects ?

Raven Herd - 5 years, 5 months ago

That sounds right. If $d(m)$ is the number of derangements of $m$ objects then there would be $\large (d(m))^{n}$ " $n$ -derangements" of the $n$ sets each of order $m$ .

Brian Charlesworth - 5 years, 5 months ago

Hmmm...... Mathematics is damn too symmetrical .

Raven Herd - 5 years, 5 months ago

@Raven Herd – Hahaha Indeed. :)

Brian Charlesworth - 5 years, 5 months ago

I got here from the page on Double Counting . How are these two related?

Shubhamkar Ayare - 2 years, 8 months ago

7.7.7 -3.3.3.3.3=100

abraham ovienloba - 1 year, 3 months ago

sorry wrong solution

abraham ovienloba - 1 year, 3 months ago

Sweets and Chocolates!

Image Credit: Wikimedia Candies

The answer is 3437316.

1 solution

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