choosing a mayor

[mule]

Hi, Our city mayor recently resigned. It was up to the remaining four council members to select a new mayor to fill out the remaining term. There were six canidates who cast their hat into the ring. Two of the four councilmen selected three of the six canidates in the same order. What is the probability of that happening by chance?
The four sitting council members were asked by the city administrator to choose three canidates in order of preference.

 

[soroban]

Hello, mule!

The problem is not clearly stated.
We need more information . . .

Our city mayor recently resigned.
It was up to the remaining four council members to select a new mayor to fill out the remaining term.
There were six canidates who cast their hat into the ring.
Two of the four councilmen selected three of the six canidates in the same order.
What is the probability of that happening by chance?

This was not explained:
. . Each councilman votes for three of the candidates?
. . And places them in order of preference?

Then how is the new mayor selected?

 

[chrisr]

Number of arrangements of 3 from 6 is [sup:3r214t6a]6[/sup:3r214t6a]P[sub:3r214t6a]3[/sub:3r214t6a]

For 2 councilmen, any choice of order for 1 man can be paired with any choice of order for the 2nd..

That gives [sup:3r214t6a]6[/sup:3r214t6a]P[sub:3r214t6a]3[/sub:3r214t6a]([sup:3r214t6a]6[/sup:3r214t6a]P[sub:3r214t6a]3)[/sub:3r214t6a] possible groupings of choices for 2 councilmen.

There are [sup:3r214t6a]4[/sup:3r214t6a]C[sub:3r214t6a]2[/sub:3r214t6a] groups of 2 councilmen.

Therefore there are [sup:3r214t6a]4[/sup:3r214t6a]C[sub:3r214t6a]2[/sub:3r214t6a]([sup:3r214t6a]6[/sup:3r214t6a]P[sub:3r214t6a]3[/sub:3r214t6a])([sup:3r214t6a]6[/sup:3r214t6a]P[sub:3r214t6a]3[/sub:3r214t6a]) possible choice pair-ups (matching and non-matching).

For each pair of men, there are [sup:3r214t6a]6[/sup:3r214t6a]P[sub:3r214t6a]3[/sub:3r214t6a] possible ways of choosing the same order.

Therefore the probability of 2 councilment picking the same 3 candidates in the same order is \(\displaystyle \frac{4c_2(6p_3)}{4c_2(6p_3)^2}\)

\(\displaystyle =\frac{1}{6p_3}=\frac{1}{6(5)4}=\frac{1}{120}\)