- Jika n! = nx(n-1)x(n-2)x … x1
untuk setiap bilangan bulat positif n, dan C(a,
b) = a!/(b!(a–b)!), untuk
a > b, keduanya bilangan bulat positif.
Berapakah C(7, 3) x C(4, 2) x C(2,1) x C(1,1) ?
- 1240
- 420
- 33452
- 115420
- 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
\(\mathrm{C}_{1}^{1} = 1\)