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### An election, with a twins twist

A city council consists of 10 members, from whom will be selected 4 officers: President, Vice-President, Secretary, and Ombudsman. But two of the members of the council are identical twins, and no one can tell them apart! If the twins are "interchangeable", whether elected to an office or not, in how many ways can the positions be filled ?

(Historical note: In 2006-2007, the identical twin Kaczynski brothers were President and Prime Minister of Poland.)

#### Normal straightforward solution:

Case 0:
If neither twin is elected, we have 8 * 7 * 6 * 5 = 1680 arrangements.

Case 1:
If one twin is elected, we have 4 positions for the twin * 8 * 7 * 6 ways to fill the other positions = 1344 arrangements.

Case 2:
If both twins are elected, we have 4C2 = 6 choices of positions for them * 8 * 7 = 6 * 56 = 336 arrangements.

Total of all three cases: 3360.

#### More interesting solution:

Total, unrestricted arrangements = 10 * 9 * 8 * 7 = 5040.
Since 1680 have neither twin (as found at left), that leaves 3360.

For each of those, write the names of those elected
in order by office. If only one twin is elected, add the other's name
as a fifth name, so both are always listed.

Suppose the twins are named Alice and Beth.
In half the arrangements, Alice's name is first,
and in the other half, Beth's name is first.

Since it doesn't matter which one comes first
(for either of two offices, or one elected and the other not),
toss out the half where Beth's name is before Alice's.

That leaves 3360 / 2 = 1680 for these cases.

The total then is 1680 "no twin" cases + 1680 "one or two twin" cases,
and again we have the total of 3360.