Probability : permutation or choose

Question

A club has $18$ members.

(a) How many ways are there to choose four members of the club to serve on an executive committee? Solution: $\binom{18}{4} = 3060$

(b) How many ways are there to choose a president, vice president, secretary, and treasurer of the club, where no person can hold more than one office? Solution: $P(18,4)=18 \cdot 17 \cdot 16 \cdot 15=73,440$

• Sometimes, I am confused when to pick up the permutation (P) or the choose. Correct me if i am wrong but,

• When the order matters $\to$ choose

• When the order does not matter $\to$ permutation

Answer 1

(a) How many ways are there to choose four members of the club to serve on an executive committee?

Solution: $\binom {18}4=3060$

You are selecting four from eigteen members.

If you are counting ways to select items, that is a count of combinations.   Combinations are not distinguished by ordering. (Order is not important.)

• $\binom n r$ or $^n\mathrm C_r$ counts combinations of $r$ elements selected from a set of $n$ $${^n\mathrm C_r}= \dfrac{n!}{r!\cdot (n-r)!}$$

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(b) How many ways are there to choose a president, vice president, secretary, and treasurer of the club, where no person can hold more than one office?

Solution: $P(18,4)=18⋅17⋅16⋅15=73,440$

You are selecting and arranging four from eighteen members into four specific positions.

If you are counting ways to select and arrange items, that is a count of permutations.   Permutations are distinguished by ordering. (Order is important!)

• $P(n,r)$ or $^n\mathrm P_r$ counts permutations of arrangements for $r$ elements selected from a set of $n$.

  $${^n\mathrm P_r}= \dfrac{n!}{(n-r)!}$$

• ${^n\mathrm P_r}={^n\mathrm C_r}\cdot r!$