5.2 - Arrangements with restrictions

5.2 - Arrangements with restrictions

An arrangement count changes when some objects look identical or some positions are restricted. In this part of 5.2, you will remove duplicate arrangements, keep specified objects together or apart, and count seating in several rows by treating each seat as a distinct position.

Repeated objects

How many different arrangements can you make from the letters of BANANA? Writing 6!6! treats the three As and two Ns as different objects. Swapping two As does not produce a new visible word, so that count is too large.

Imagine temporarily labelling the As as A1,A2,A3A_1,A_2,A_3 and the Ns as N1,N2N_1,N_2. Each visible arrangement comes from 3!3! orders of the labelled As and 2!2! orders of the labelled Ns. It has been counted 3!2!3!2! times.

Arrangements with identical objects

If nn objects contain groups of a,b,…a,b,\ldots identical objects, with different groups distinguishable, the number of different linear arrangements is

n!a!b!⋯.\frac{n!}{a!b!\cdots}.

Include a factorial for each repeated group. The group sizes, including single objects, must account for all nn positions.
Not in the formula booklet: learn the overcounting argument.

The division is justified by indistinguishability. Two different people who share an initial are still distinct people, so their seating arrangements would not be divided by 2!2!.

Worked example 1: two repeated groups

Question. Find the number of different arrangements of the six letters of BANANA.

Plan. Count six labelled positions, then remove permutations of the three identical As and two identical Ns.

6!3!2!=60.\frac{6!}{3!2!}=60.

B occurs once, so its optional denominator factor is 1!=11!=1.

Answer. 6060 arrangements.

Check. Choose three of the six positions for A, then two of the remaining three for N; B fills the last position. This gives (63)(32)=20×3=60\binom63\binom32=20\times3=60.

Add one new distinct letter, D, to make BANDANA. The sizes of the repeated groups remain unchanged.

Worked example 2: one extra distinct object

Question. Find the number of different arrangements of the seven letters of BANDANA.

Plan. Use seven positions, still dividing only for the three As and two Ns.

7!3!2!=420.\frac{7!}{3!2!}=420.

Answer. 420420 arrangements.

Check. Insert D into any of the seven gaps around the six letters of an arrangement from Example 1. Removing the unique D reverses the process, so 60×7=42060\times7=420.

The opening question is answered by removing exactly the orders that nobody can distinguish. It is not a rule to divide by every factorial visible in a question.

Watch out

For BANANA, 6!/(3!+2!)6!/(3!+2!) is incorrect. The A orders and N orders vary together, so their repeated counts multiply: divide by 3!2!3!2!.

Check the repeated groups before you calculate.

Together or apart

To keep two specified people adjacent, temporarily join them into one block. The block takes one place in the outside order, but the people can still exchange positions inside it.

For six distinct people including A and B, the five outside units are the block AB and the other four people. Each outside arrangement has two internal orders, AB or BA. For three specified distinct people kept together, there would instead be 3!3! internal orders.

A block

A block is a group treated as one unit while arranging the units around it. Count its permitted internal arrangements separately. “A and B are together” means adjacent here; “A and B are in the same row” need not mean adjacent.

If the restriction is “not adjacent”, subtract the adjacent arrangements from all arrangements. The two sets form a complete, disjoint split.

Worked example 3: two people together

Question. Six distinct people A, B, C, D, E and F stand in a line. Find the number of arrangements in which A and B stand next to each other.

Plan. Arrange the AB block with the other four people, then count the two orders within the block.

5!×2!=120×2=240.5!\times2!=120\times2=240.

Answer. 240240 arrangements.

Check. The pair can occupy any of five adjacent position pairs. Its two orders and the 4!4! orders of everyone else give 5×2×4!=2405\times2\times4!=240.

Change only the adjacency condition. Rather than trying to build every possible gap pattern, remove the forbidden cases already counted.

Worked example 4: the same two people apart

Question. The same six people stand in a line. Find the number of arrangements in which A and B do not stand next to each other.

Plan. Subtract all adjacent arrangements from the unrestricted 6!6! arrangements.

6!−2(5!)=720−240=480.6!-2(5!)=720-240=480.

Answer. 480480 arrangements.

Check. A and B can occupy (62)−5=10\binom62-5=10 non-adjacent position pairs. Assign A and B to them in two ways, then arrange the other four: 10×2×4!=48010\times2\times4!=480.

Blocks also combine with repeated objects. Suppose the symbols A, A, B, B, B, C, D are arranged with A at both ends. The five middle symbols have 5!/3!=205!/3!=20 arrangements. Exactly 3!=63!=6 have all three Bs together, because the outside units are BBB, C and D. Thus 20−6=1420-6=14 have the As at the ends and the three Bs not all together. “Not all together” allows a pair of adjacent Bs; it does not require every B to be separated. Two identical As at the two ends have no extra 2!2! ordering factor.

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