Soal dan Pembahasan OSK Komputer 2007

  1. Jika n! = nx(n-1)x(n-2)x … x1 untuk setiap bilangan bulat positif n, dan C(a, b) = a!/(b!(ab)!), untuk a > b, keduanya bilangan bulat positif. Berapakah C(7, 3) x C(4, 2) x C(2,1) x C(1,1) ?
      1. 1240
      2. 420
      3. 33452
      4. 115420
      5. 22

    Sudah diketahui pada soal bahwa
    \(\mathrm{C}_{b}^{a} = \frac{a!}{b!(a-b)!}\)
    Sehingga hanya cukup dihitung masing-masing bagian.

    \(\mathrm{C}_{7}^{3} = \frac{a!}{b!(a-b)!}\)
    \(\mathrm{C}_{7}^{3} = \frac{7!}{3!(7-3)!}\)
    \(\mathrm{C}_{7}^{3} = \frac{7 \times 6 \times 5 \times 4!}{3 \times 2 \times 1 \times4!}\)
    dengan menyederhanakannya menjadi
    \(\mathrm{C}_{7}^{3} = \frac{7 \times 6 \times 5}{6}\)
    \(\mathrm{C}_{7}^{3} = 7 \times 5\)
    \(\mathrm{C}_{7}^{3} = 35\)

    \(\mathrm{C}_{4}^{2} = \frac{4!}{2!(4-2)!}\)
    \(\mathrm{C}_{4}^{2} = \frac{4 \times 3 \times 2!}{2\times 1 \times 2!}\)
    \(\mathrm{C}_{4}^{2} = 2 \times 3\)
    \(\mathrm{C}_{4}^{2} = 6\)

    \(\mathrm{C}_{2}^{1} = \frac{2!}{1!(2-1)!}\)
    \(\mathrm{C}_{2}^{1} = \frac{2 \times 1}{1 \times 1}\)
    \(\mathrm{C}_{2}^{1} = 2\)

    \(\mathrm{C}_{1}^{1} = \frac{1!}{1!(1-1)!}\)
    \(\mathrm{C}_{1}^{1} = \frac{1!}{1!(0!)}\)
    catatan : 0! = 1
    \(\mathrm{C}_{1}^{1} = 1\)

    Sehingga C(7, 3) x C(4, 2) x C(2,1) x C(1,1) = 35 x 6 x 2 x 1 = 420 B.

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