# Eight first class and six second class petty officers .?

• December 2nd 2009, 04:18 AM
aeroflix
Eight first class and six second class petty officers .?
Eight first class and six second class petty officers are on the board of the 56 club. In how many ways can the members elect, from the board, a president, a vice-president, a secretary, and a treasurer if the president and secretary must be first class petty officers and the vice-president and treasurer must be second class petty officers?

also.. the second question is

For the preceding example, suppose we are asked the following: In how many ways can the members elect the office holders from the board if two of the office holders must be first class petty officers and two of the office holders must be second class petty officers?
• December 2nd 2009, 06:00 AM
Soroban
Hello, aeroflix!

Here's the first part . . .

Quote:

Eight PO1's and six PO2's are on the board of the club.
In how many ways can the members elect, from the board,
a president, a vice-president, a secretary, and a treasurer
if the president and secretary must be PO1's
and the vice-president and treasurer must be PO2's?

We choose a President and a Secretary from the 8 PO1's.
. . There are: . $_8P_2 = 56$ ways.

We choose a V.P. and Treasurer from the 6 PO2's.
. . There are: . $_6P_2 = 30$ ways.

Therefore, there are: . $56\cdot30 \,=\,1680$ ways.

• December 2nd 2009, 12:20 PM
aeroflix
hello, thanks,, i dont get the second part.
• December 2nd 2009, 03:59 PM
emakarov
For the second question, I think you first need to select two offices (positions) out of four. They will be taken by first-class officers. The remaining two will be taken by second-class officers, so there is no choice in selecting the remaining positions. After that, you proceed as in question 1.
• December 2nd 2009, 10:00 PM
aeroflix
so i will mutiply the answer to 4C2? why?? cant it be 4C2 x 4C2, one for first class and one for second class
• December 3rd 2009, 02:23 AM
emakarov
Quote:

so i will mutiply the answer to 4C2?
Yes.
Quote:

cant it be 4C2 x 4C2, one for first class and one for second class
To count the number of ways to select something you need to represent the selection procedure as a series of choices. For example, to express $P(n,k)$ through $C(n,k)$ (i.e., the number of ordered sequences of length $k$ vs. the number of unordered sets of the same length) one can do the following. To choose a sequence of length $k$ from a set of $n$ elements, one first selects an unordered set of $k$ elements and then selects an ordering of those $k$ elements. So $P(n,k)=C(n,k)\cdot k!$.

The description above is a simplification because sometimes you need to break the set of all possible variants into disjoint subsets, and then proceed as described.

The important thing about the sequence of choices is that they (1) allow one to generate only those variants that need to be counted, (2) allow one to come up with all possible such variants, and (3) do not produce the same variant twice. Let's look at the number of ways to select which of the 4 positions are taken by the first class and which are taken by the second class (without looking into how first-class positions are distributed). I claim that there is only one choice: select any set $\{a, b\}$ where $a,b\in\{1,2,3,4\}$. After this choice, positions $\{a, b\}$ go to the first class and positions $\{1,2,3,4\}-\{a,b\}$ go to the second class. If I am not mistaken, we generated all variants and did not produce the same variant twice.

On the other hand, your version $C(4,2)\cdot C(4,2)$ implies two choices. Presumably, during the first choice we select two positions for the first class. Then, without removing those selected positions from the set $\{1,2,3,4\}$, we select two positions for the second class. This way, for example, we can come up with the set $\{1,2\}$ for the first class and $\{1,3\}$ for the second class. But $\{1,2\}$ and $\{1,3\}$ is not a legal variant.
• December 5th 2009, 07:53 AM
aeroflix
thanks very much!!!

great help