赤玉5個、黒玉3個、白玉2個を(4個, 3個, 3個)に分ける方法（もらう人を区別しない）」を全て列挙せよ

library(gtools) # permutations を使用

# 各色の玉の個数
balls <- c(赤 = 4, 黒 = 3, 白 = 2, 青 = 1)

# 各箱のサイズ
box_sizes <- c(4, 3, 3)

# 指定したサイズで玉を選ぶ関数
generate_combinations <- function(balls, size) {
colors <- names(balls)
all_combinations <- permutations(n = length(colors), r = size, v = colors, repeats.allowed = TRUE)

# 各組み合わせが元の玉の数を超えないようにフィルタリング
valid_combinations <- all_combinations[apply(all_combinations, 1, function(combo) {
all(table(combo) <= balls[names(table(combo))])
}), ]

return(unique(t(apply(valid_combinations, 1, sort)))) # 各組み合わせをソートして重複を削除
}

# すべての分割を求める
unique_distributions <- list()

first_box_choices <- generate_combinations(balls, box_sizes[1])

for (first_box in 1:nrow(first_box_choices)) {
remaining1 <- balls
first_selection <- first_box_choices[first_box, ]

for (color in first_selection) {
remaining1[color] <- remaining1[color] - 1
}

second_box_choices <- generate_combinations(remaining1, box_sizes[2])

for (second_box in 1:nrow(second_box_choices)) {
remaining2 <- remaining1
second_selection <- second_box_choices[second_box, ]

for (color in second_selection) {
remaining2[color] <- remaining2[color] - 1
}

# 第3箱は残りの玉で確定
third_box <- rep(names(remaining2), times = remaining2)

# 各箱内をソートし、3つの箱の並びも統一
sorted_distribution <- list(sort(first_selection), sort(second_selection), sort(third_box))
sorted_distribution <- sorted_distribution[order(sapply(sorted_distribution, paste, collapse = ""))] # 箱順も統一

# **文字列化してリストを扱いやすくする**
unique_distributions <- append(unique_distributions, list(paste(sapply(sorted_distribution, paste, collapse = ","), collapse = " | ")))
}
}

# 重複を削除
unique_distributions <- unique(unique_distributions)

# 総数を確認（正しく63になるはず）
length(unique_distributions)

# 結果を表示
print(unique_distributions)