AMC 10 · 2013 · #11

Grade 7 counting

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search

combinations-basicquadratic-equations convert-to-algebraguess-and-check ↑ Prerequisites: combinations-basic

📏 Medium solution 💡 2 insights

Problem

A student council has some number of members. Choosing a two-person welcoming committee can be done in exactly 10 ways. Students may serve on both committees, so the same members are available for the planning committee. Find how many different three-person planning committees can be chosen from the same council.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Introduce a Variable

The council size is hidden. Name it with a variable, use the 10 ways to pick a pair to pin down how many members there are, then count the three-person committees for that many members.

1STEP 1

Name the number of members

Let n be the number of students on the council; a welcoming committee is any 2 of them, order ignored.

n = number of council members

2STEP 2

Count 2-person groups

Ordered picks give n(n-1), but each pair is counted twice, so there are n(n-1)/2 two-person committees.

C(n, 2)=(n(n-1))/2

3STEP 3

Solve for n

So n(n-1)/2 = 10 gives n(n-1) = 20, and the only consecutive pair with product 20 is 5 and 4, so n = 5.

(n(n-1))/2=10 → n(n-1)=20 → 5 · 4=20 → n=5

4STEP 4

Count 3-person groups

Choosing 3 of 5 in order gives 5 · 4 · 3 = 60, and each trio is counted 6 times, so 60/6 = 10 committees, answer (A).

C(5, 3)=(5 · 4 · 3)/(3 · 2 · 1)=60/6=10

Answer

Check n = 5 against the first clue: choosing 2 of 5 gives (5 · 4)/2 = 10 pairs, which matches. Then choosing 3 of 5 also gives 10. A quick symmetry check confirms it: choosing 3 people to include is the same as choosing the 2 people to leave out, and there are 10 ways to leave out a pair. So the count is 10, answer (A).

💡Key takeaway

When a count is given to you, work backwards to find how many things there are, then count the new way.

Name the number of members
Count 2-person groups
Solve for n
Count 3-person groups

Name the number of members

Count 2-person groups

Solve for n

Count 3-person groups

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