In how many ways can seven friends be seated in a row at the theatre?

Question

BlackTom AI · Accepted Answer

The problem asks for the number of ways seven distinct friends can be seated in a row.
Option a: 7! / 2! suggests dividing by 2 for some symmetry or indistinguishability, which applies to problems where two items are considered identical or where only relative order matters in a reduced set. Here, all seven friends are distinct and each seating arrangement is different, so dividing by 2! is inappropriate.
Option b: 7! represents the number of permutations of seven distinct items, which exactly matches the scenario of seating seven different friends in a row where the order matters. Each of the seven positions can be filled by one of the remaining friends, giving 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040 arrangements.
Option c: 7! / (2!5!) equals 21, which corresponds to the number of ways to choose 2 items from 7 (a binomial coefficient) and does not account for the ordering of all seven individuals. It combines a factorial with a division that removes order among two and five groups, not appropriate for full row seating of distinct people.
Option d: 7 is far too small; it might reflect choosing a single seat from seven or counting a trivial case, but it ignores the orders of the remaining six seats and does not reflect permutation counts.
Option e: 7C1 equals 7, which again is a binomial coefficient for choosing one item from seven, not the total number of permutations of seven distinct people in a line. It omits all ordering information and is not applicable to this seating problem.

In how many ways can seven friends be seated in a row at the theatre?

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