This question's explanation and your answers so far.
所属するグループ又はメンバに設定した属性情報によって，人事ファイルへのアクセス権を管理するシステムがある。人事部グループと，所属するメンバA～Dの属性情報が次のように設定されているとき，人事ファイルを参照可能な人数と更新可能な人数の組合せはどれか。 〔属性情報の設定方式〕 (1) 属性情報は3ビットで表される。 (2) 各ビットは，左から順に参照，更新，削除に対応し，1が許可，0が禁止を意味する。 (3) グループの属性情報は，メンバの属性情報が未設定の場合にだけ適用される。 〔属性情報の設定内容〕 人事部グループ：100 メンバA：100，メンバB：111，メンバC：110，メンバD：未設定
Reference 3, update 1. For reference, all four of A, B, C, and D have a leftmost bit of 1, so it is 4 people, not 3; and update is B and C, 2 people, not 1. This option is wrong on both counts.
Reference 3, update 2. The number who can update (2) is correct, but treating the number who can reference as 3 is wrong. Because D is unset, the group's 100 applies and reference is allowed, so 4 people can reference.
Reference 4, update 1. The number who can reference (4) is correct, but treating the number who can update as 1 is wrong. Update is B and C, whose middle bit is 1, making 2 people, not 1.
正解 The attribute information consists of 3 bits corresponding, from left to right, to reference, update, and delete, where 1 means allowed. By rule (3), the group value 100 applies only when a member's attribute is unset. Each person's effective attribute is: A is 100, B is 111, C is 110, and D is unset so the group's 100 applies. Reference is allowed for those whose leftmost bit is 1; since A, B, C, and D are all 1, 4 people can reference. Update is allowed for those whose middle bit is 1; only B's 1 and C's 1 qualify (A and D are 0), so 2 people can update. Therefore the combination is reference 4 and update 2, so エ is correct.