問63
頻出度★★☆
a, b, c, d, e, fの6文字を任意の順で1列に並べたとき, aとbが隣同士になる場合は, 何通りか。
平成26年度 春期 ITパスポート試験 問63
テクノロジ系基礎理論 › 応用数学
This question's explanation and your answers so far.
解説
問 63
Incorrectア
120 is the value you get if you bundle a and b into a block, compute the number of arrangements of the 5 elements as 5! = 120, and stop the calculation there. The inside of the block still has two orderings, “ab” and “ba”, so you must finally multiply by 2! = 2. What is missed here is that handling an adjacency condition by blocking takes two steps: the arrangements after bundling × the arrangements inside the block.
- イ
正解 The correct answer is イ, 240 arrangements. The condition that two letters are adjacent is handled with the standard technique of bundling the adjacent pair into a single unit (a block). First, treat a and b as one block “ab”. Then the items to be arranged are the “ab” block, c, d, e, and f — five items in all. The number of ways to arrange these five in a row is 5! = 5 × 4 × 3 × 2 × 1 = 120. Next, do not forget that the inside of the block also has orderings. The order of a and b can be “ab” or “ba”, two ways, so multiply by the internal arrangements 2! = 2. Therefore the number of arrangements we want is 120 × 2 = 240, and イ is correct. As a check: the total number of arrangements of all 6 letters is 6! = 720, and the cases where a and b are adjacent make up 240 ÷ 720 = 1/3 of them. Among the 6 positions, the adjacent position pairs are (1,2)(2,3)(3,4)(4,5)(5,6), that is 5 pairs, while the total number of position pairs that a and b can occupy is 6 × 5 ÷ 2 = 15, so the probability of being adjacent is 5/15 = 1/3. This agrees with the check above.
Incorrectウ
720 is 6! = 720, that is, the total number of arrangements of all 6 letters without asking whether a and b are adjacent. It narrows nothing down by the condition, so it is wrong. In fact, the required 240 arrangements are exactly one third of these 720.
Incorrectエ
1,440 is 6! × 2 = 1,440, the value you get by taking the overall 720 arrangements and multiplying them again by the 2 orderings inside the block. After blocking, what you must count is the arrangements of 5 elements, not 6, that is 5!; using 6! means doubling without first removing the duplication that bundling into a block introduced. Besides, a count restricted by a condition can never exceed the overall total of 720, so you can notice the error at this point alone.
ここがポイント
- Total number of permutations (arrangements): the number of ways to arrange n distinct items in a row is n! = n × (n−1) × … × 2 × 1. In this question, all 6 letters give 6! = 720, while the 5 elements after blocking give 5! = 120.
- The standard technique for an adjacency condition is to treat the adjacent elements as one bundled block. The procedure has two steps: (1) regard the block as a single element and arrange the whole set, and (2) multiply by the ordering inside the block. In this question that gives 5! × 2! = 240. Forgetting to multiply by the internal arrangements, and miscounting the number of elements after blocking (treating 5 as 6), are the typical ways points are lost.
用語