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Permutation & combination · Lesson 2 of 2

Committees with conditions and probability

Committees with 'at least', 'at most', a particular person in or out, and people who can't serve together; splitting people into groups; and probabilities found by counting selections.

22 minYou should already know: Counting
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A committee is a selection, so it is counted with nCr{}^nC_r (see counting). When it is chosen from two groups, choose from each group and multiply. The conditions in a question decide which splits count.

Splitting into cases

“At least” and “at most” allow several splits. List each split that fits, count it, then add:

3 people, at least 2 women (4 women, 5 men)2 W + 1 M⁴C₂ × ⁵C₁ = 30+3 W⁴C₃ = 430 + 4 = 34 committeeslist the cases that fit, count each, then add
At least 2 womenCount each case, then add

When the unwanted cases are fewer, use the complement: all committees minus the ones that break the condition. “At least one” is almost always quicker this way.

Committees with conditionsPick a condition
menwomenways
40⁶C₄ × ⁵C₀ = 15 × 1 = 15
31⁶C₃ × ⁵C₁ = 20 × 5 = 100
22⁶C₂ × ⁵C₂ = 15 × 10 = 150
13⁶C₁ × ⁵C₃ = 6 × 10 = 60
04⁶C₀ × ⁵C₄ = 1 × 5 = 5
215committees that fit0.6515probability, out of 330
At least 2 means 2, 3 or 4 women: add those cases. Chosen at random, the probability is 215 ÷ 330 ≈ 0.6515.

Worked example · WAEC 2011

WAEC 2011 · Paper 2 · Q13 (a)

A committee of five is to be formed among 6 Ghanaians, 8 Nigerians and 5 Gambians. In how many ways can the committee be formed if: (i) there is no restriction; (ii) at most 2 Ghanaians are on the committee; (iii) 1 Nigerian is on the committee?

  1. No restriction

    • There are 6+8+5=19{6 + 8 + 5 = 19} people.
    • So there are 19C5=11 628{{}^{19}C_5 = 11\,628} committees.

    Think first. Choose 5 from how many people?

  2. At most 2 Ghanaians

    • 0 Ghanaians: 6C0×13C5=1×1287=1287{{}^6C_0 \times {}^{13}C_5 = 1 \times 1287 = 1287}.
    • 1 Ghanaian: 6C1×13C4=6×715=4290{{}^6C_1 \times {}^{13}C_4 = 6 \times 715 = 4290}.
    • 2 Ghanaians: 6C2×13C3=15×286=4290{{}^6C_2 \times {}^{13}C_3 = 15 \times 286 = 4290}.
    • Add: 1287+4290+4290=9867{1287 + 4290 + 4290 = 9867}.

    Think first. Which numbers of Ghanaians are allowed? The rest come from the 13 others.

  3. Exactly 1 Nigerian

    • 1 Nigerian from 8: 8C1=8{{}^8C_1 = 8}.
    • The other 4 from the 11 non-Nigerians: 11C4=330{{}^{11}C_4 = 330}.
    • So 8×330=2640{8 \times 330 = 2640}.

    Think first. 1 from the 8 Nigerians, and the other 4 from whom?

A particular person in or out

  • A particular person must be on the committee: put them on, then choose the rest from everyone else.
  • A particular person must not be on it: remove them, then choose from the rest.
  • Two people cannot serve together: count all committees, then take away the ones with both.

Worked example · WAEC 2018

WAEC 2018 · Paper 2 · Q12 (b)

A committee of three men and two women is to be formed from four men and six women. How many different committees can be formed if: (i) there are no restrictions; (ii) a particular man and a particular woman cannot serve together on the same committee?

  1. No restriction

    • 4C3×6C2=4×15=60{{}^4C_3 \times {}^6C_2 = 4 \times 15 = 60} committees.

    Think first. 3 men from 4 and 2 women from 6.

  2. Both of them together

    • The other 2 men from 3: 3C2=3{{}^3C_2 = 3}.
    • The other woman from 5: 5C1=5{{}^5C_1 = 5}.
    • So 3×5=15{3 \times 5 = 15} committees have both of them.

    Think first. Put the particular man and woman on. Choose the rest.

  3. Not together

    • 60−15=45{60 - 15 = 45} committees.

More: committees and selections

Splitting people into groups

To split nn people into groups of given sizes, divide n!n! by the factorial of each group size, just as for repeated letters. If kk of the groups are the same size and have no names, swapping those groups gives the same split, so divide by k!k! as well. For n=mkn = mk people in kk unnamed groups of mm:

ways=n!(m!)k k!\text{ways} = \frac{n!}{(m!)^k \, k!}
pair 1, pair 2same splitABCD=CDABACBD=BDACADBC=BCAD4! ÷ (2! 2!) = 6 numbered, ÷ 2! = 3 splits
Unnamed pairsA, B, C, D into two pairs: numbering the pairs counts each split 2! times
  • Groups of different sizes can’t be swapped, so there is no k!k!: 9 people into groups of 2, 3 and 4 is 9!2! 3! 4!\frac{9!}{2!\,3!\,4!}.
  • Groups with names (team A and team B, room 1 and room 2) can’t be swapped either: no k!k!.

Worked example

In how many ways can 8 players be split into 4 pairs for a doubles tournament?

  1. Line them up and cut

    • Order all 8 players: 8!=40 320{8! = 40\,320} ways.
    • Cut the line into pairs: the first two, the next two, and so on.
    • Swapping the two in a pair changes nothing: divide by (2!)4=16{(2!)^4 = 16}.
    • That gives 40 32016=2520{\frac{40\,320}{16} = 2520} numbered splits.

    Think first. If the pairs were numbered 1 to 4, how many ways?

  2. The pairs have no names

    • The 4 pairs can be put in 4!=24{4! = 24} orders, all giving the same split.
    • So 252024=105{\frac{2520}{24} = 105} ways.

    Think first. How many times is each split counted among the 2520?

  3. Check another way

    • The first player’s partner: 7 choices.
    • The next unpaired player’s partner: 5 choices, then 3, then 1.
    • 7×5×3×1=105{7 \times 5 \times 3 \times 1 = 105}, which agrees.

    Think first. The first player picks a partner. How many choices? Then the next unpaired player?

More: splitting into groups

Probability by counting

When every selection is equally likely,

P(event)=number of selections that fittotal number of selectionsP(\text{event}) = \frac{\text{number of selections that fit}}{\text{total number of selections}}

Count both with nCr{}^nC_r, using the same method as for committees.

Worked example · WAEC 2017

WAEC 2017 · Paper 2 · Q5

A committee of 5 members is to be formed from 6 men and 7 women. Calculate the probability that it consists of:

all women;

2 men and 3 women.

  1. All committees

    • 13C5=1287{{}^{13}C_5 = 1287}.

    Think first. Choose 5 from 13.

  2. All women

    • 7C5=21{{}^7C_5 = 21}.
    • P=211287=7429≈0.0163{P = \frac{21}{1287} = \frac{7}{429} \approx 0.0163}.

    Think first. Choose all 5 from the 7 women.

  3. 2 men and 3 women

    • 6C2×7C3=15×35=525{{}^6C_2 \times {}^7C_3 = 15 \times 35 = 525}.
    • P=5251287=175429≈0.4079{P = \frac{525}{1287} = \frac{175}{429} \approx 0.4079}.

More: probability by counting

Your turn

WAEC 2011 · Paper 2 · Q6

  1. (a)

    Evaluate 9P315C3+5C33P2\dfrac{{}^9P_3}{{}^{15}C_3} + \dfrac{{}^5C_3}{{}^3P_2}, correct to two decimal places.

  2. (b)

    A committee of 2 tutors and 5 pupils is to be formed among 6 tutors and 10 pupils. In how many ways can this be done, if one particular tutor must be on the committee and two particular pupils must not be on the committee?

Worked solution (try it first)

(a)

  1. Work out each symbol:  9P3=9×8×7=504\,{}^9P_3 = 9 \times 8 \times 7 = 504 and  15C3=15×14×136\,{}^{15}C_3 = \dfrac{15 \times 14 \times 13}{6}
    =455= 455.
  2.  5C3=10\,{}^5C_3 = 10 and  3P2=3×2=6\,{}^3P_2 = 3 \times 2 = 6.
  3. So the value is 504455+106=1.1077+1.6667\dfrac{504}{455} + \dfrac{10}{6} = 1.1077 + 1.6667
    =2.7744= 2.7744, which is 2.772.77 to two decimal places.

(b)

  1. Put the particular tutor on.
  2. Choose the second tutor from the other 5:  5C1=5\,{}^5C_1 = 5.
  3. Leave out the two particular pupils.
  4. Choose 5 pupils from the other 8:  8C5=56\,{}^8C_5 = 56.
  5. So 5×56=2805 \times 56 = 280 ways.

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