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WAEC 2008 · Paper 2 · Q14✱

  1. (a)

    (i) In how many ways can eight boys be seated in a row? (ii) If two of the boys in (a)(i) cannot sit together, in how many ways can they be seated?

    Separate values with commas, e.g. 3, −2

  2. (b)

    A delegation of a labour union consists of 7 men and 3 women. If 4 of them are selected at random to give a talk, what is the probability of selecting: (i) 3 men and 1 woman; (ii) at least 2 women?

    Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)(i)

  1. Eight boys in a row: 8!=40 3208! = 40\,320 ways.

(ii)

  1. First count the ways with the two boys together: treat them as one unit, giving 7!7! arrangements, and they can swap places in 22 ways.
  2. Together: 2×7!=10 0802 \times 7! = 10\,080 ways.
  3. Not together: 8!−2×7!=40 320−10 0808! - 2 \times 7! = 40\,320 - 10\,080
    =30 240= 30\,240 ways.

(b)

  1. Number of ways to choose any 4 of the 10: 10C4=210^{10}C_4 = 210.

(i)

  1. 3 men and 1 woman: 7C3×3C1=35×3=105^7C_3 \times {^3C_1} = 35 \times 3 = 105.
  2. Probability =105210=12= \frac{105}{210} = \frac12.

(ii)

  1. At least 2 women means 2 or 3 women. 2 women: 7C2×3C2=21×3=63^7C_2 \times {^3C_2} = 21 \times 3 = 63. 3 women: 7C1×3C3=7^7C_1 \times {^3C_3} = 7.
  2. Probability =63+7210= \frac{63 + 7}{210}
    =70210= \frac{70}{210}
    =13= \frac13.

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