Satu paket siomay terdiri dari $10$ buah, yakni dapat terdiri dari siomay, tahu, telur, kentang, kol, dan pare. Jika tidak suka pare atau kol, bisa diganti dengan bahan lain. Jumlah pare dan kol maksimal $1$ buah. Telur harus $1$ buah dan siomay minimal $3$ buah. Banyak cara pemilihan isi satu paket Siomay adalah..

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Q. Satu paket siomay terdiri dari

10

buah, yakni dapat terdiri dari siomay, tahu, telur, kentang, kol, dan pare. Jika tidak suka pare atau kol, bisa diganti dengan bahan lain. Jumlah pare dan kol maksimal

1

buah. Telur harus

1

buah dan siomay minimal

3

buah. Banyak cara pemilihan isi satu paket Siomay adalah..

Constraints: We need to determine the number of ways to fill a package of Siomay with $10$ items, given the constraints. The constraints are: $1$ egg (telur) must be included, at least $3$ Siomay must be included, a maximum of $1$ Pare and $1$ Kol can be included, and Pare or Kol can be replaced with another item if not liked.
Mandatory Items: First, we include the mandatory items in the package. We have $1$ egg and at least $3$ Siomay, which together make $4$ items. This leaves us with $6$ more items to choose for the package.
Case $1$ : Since we can have at most $1$ Pare and $1$ Kol, we consider the cases where we have $0$ or $1$ of each. This gives us $4$ cases: $(0$ Pare, $0$ Kol), $(0$ Pare, $1$ Kol), $(1$ Pare, $0$ Kol), and $(1$ Pare, $1$ Kol).
Case $2$ : For each case, we need to determine how many ways we can choose the remaining items (Tahu, Kentang, and additional Siomay) to fill up the rest of the package.
Case $3$ : Case $1$ : $0$ Pare, $0$ Kol) We have $6$ items left to choose, and they can be any combination of Tahu, Kentang, and additional Siomay. Since there are no restrictions on these items, we use the stars and bars method to calculate the combinations. The formula is $n+k-1$ choose $k-1$ , where $n$ is the number of items to choose from ( $3$ in this case, since we can choose more Siomay, Tahu, or Kentang) and $k$ is the number of items to choose ( $6$ items). So we have $3+6-1$ choose $0$ $0$ = $0$ $1$ choose $0$ $2$ .
Case $4$ : Case $2$ : $0$ Pare, $1$ Kol) We have $5$ items left to choose, and they can be any combination of Tahu, Kentang, and additional Siomay. Using the stars and bars method, we have $3+5-1$ choose $5-1$ = $7$ choose $4$ .
Calculate Combinations: Case $3$ : $(1 \text{ Pare}, 0 \text{ Kol})$ $\newline$ This case is identical to Case $2$ , as we still have $5$ items left to choose and the same options. So we have another $7 \choose 4$ combinations.
Total Combinations: Case $4$ : ( $1$ Pare, $1$ Kol) $\newline$ We have $4$ items left to choose, and they can be any combination of Tahu, Kentang, and additional Siomay. Using the stars and bars method, we have $(3+4-1)$ choose $(4-1) = 6$ choose $3$ .
Total Combinations: Case $4$ : $(1 \text{ Pare}, 1 \text{ Kol})$ $\newline$ We have $4$ items left to choose, and they can be any combination of Tahu, Kentang, and additional Siomay. Using the stars and bars method, we have $(3+4-1)$ choose $(4-1)$ = $6$ choose $3$ .Now we calculate the combinations for each case: $\newline$ Case $1$ : $8$ choose $5$ = $56$ $\newline$ Case $2$ : $7$ choose $4$ = $4$ $1$ $\newline$ Case $3$ : $7$ choose $4$ = $4$ $1$ $\newline$ Case $4$ : $6$ choose $3$ = $4$ $7$
Total Combinations: Case $4$ : $(1 \text{ Pare}, 1 \text{ Kol})$ $\newline$ We have $4$ items left to choose, and they can be any combination of Tahu, Kentang, and additional Siomay. Using the stars and bars method, we have (3+4-1) \choose (4-1) = 6 \choose 3.Now we calculate the combinations for each case: $\newline$ Case $1$ : $8 \choose 5 = 56$ $\newline$ Case $2$ : $7 \choose 4 = 35$ $\newline$ Case $3$ : $7 \choose 4 = 35$ $\newline$ Case $4$ : $6 \choose 3 = 20$ Finally, we add up the combinations from all cases to get the total number of ways to fill the package: $\newline$ Total combinations = $56 + 35 + 35 + 20 = 146$