HOTS Zone : Kombinasi dengan Pengulangan (Combination with Repetition)

Misal a = a' + 1, b = b' + 1, c = c' + 4, dan d = d' + 1, sehingga
\(\begin{aligned} a'+1+b'+1+c'+4+d'+1&=10\\ a'+b'+c'+d'&=3 \end{aligned}\)
a' < 2

Jika a' = 0, b' + c' + d' = 3
\(C^{3+3-1}_3=10\)

Jika a' = 1, b' + c' + d' = 2
\(C^{3+2-1}_2=6\)

10 + 6 = 16

d harus ganjil. Misal a = a' + 1, b = b' + 1, c = c' + 1, dan d = 2d' + 1, sehingga
\(\begin{aligned} 2(a'+1)+2(b'+1)+2(c'+1)+2d'+1&=15\\ 2a'+2+2b'+2+2c'+2+2d'+1&=15\\ 2a'+2b'+2c'+2d'&=8\\ a'+b'+c'+d'&=4 \end{aligned}\)

\(C^{4+4-1}_4=\boxed{\boxed{35}}\)

Misal bilangannya adalah $\overline{abcd}$ dengan a ≥ 1; b, c, d ≥ 0. Misal a = a' + 1.
\(\begin{aligned} a+b+c+d&=8\\ (a'+1)+b+c+d&=8\\ a'+b+c+d&=7 \end{aligned}\)

\(\begin{aligned} C_7^{7+4-1}&=C_7^{10}\\ &=\dfrac{10!}{(10-7)!\cdot7!}\\[4pt] &=\dfrac{10\cdot\cancelto{3}{9}\cdot\cancelto{4}{8}\cdot\cancel{7!}}{\cancel{3}\cancel{2}\cdot\cancel{7!}}\\ &=\color{blue}\boxed{\boxed{\color{black}120}} \end{aligned}\)

• Jika a, b, c ≡ 1 mod 3

  \(\begin{aligned} a+b+c&=60\\ (3k+1)+(3l+1)+(3m+1)&=60\\ k+l+m&=19 \end{aligned}\)
  banyak tripel = $\binom{19+3-1}{3-1}=210$
  

• Jika a, b, c ≡ 2 mod 3

  \(\begin{aligned} a+b+c&=60\\ (3k+2)+(3l+2)+(3m+2)&=60\\ k+l+m&=18 \end{aligned}\)
  banyak tripel = $\binom{18+3-1}{3-1}=190$
  

210 + 190 = 400