Word Rearrangements with Constraints

Many word-rearrangement problems add a restriction on top of the basic count: certain letters must sit next to each other, or certain letters must never sit next to each other. This lesson develops the glue technique for togetherness and the total-minus-together technique for separation, then combines both with the multinomial formula from the previous lesson.

What you will learn

• Count arrangements where certain letters must stay together
• Count arrangements where certain letters must stay apart
• Combine grouping tricks with the multinomial formula

Keep paper and a pencil nearby. Try each problem before opening a hint, and mark sections reviewed as you work through them.

Letters that must stay together

When a problem requires two or more specific letters to sit next to each other, treat that whole cluster of letters as a single glued block. Arranging the block along with the remaining letters is then an ordinary arrangement problem with one fewer item than the total letter count, since the block counts as one unit.

After arranging the block and the other letters, multiply by the number of ways to order the letters inside the block itself, since sliding the block into a position does not fix the internal order of the letters glued inside it. If the letters glued together are all different from each other, that internal count is a plain factorial of the block's size; if some of the glued letters are identical to each other, the internal count must itself be divided by the appropriate factorial, exactly as in ordinary word rearrangements.

Letters that must stay apart

A separation requirement is usually easiest to handle with complementary counting: first find the total number of arrangements with no restriction, then subtract the number of arrangements where the two letters are adjacent, computed with the glue technique from the first section. What remains is exactly the count where the two letters are never adjacent.

This works because every arrangement of the letters falls into exactly one of two groups: the two letters are adjacent, or they are not. Counting the first group and subtracting it from the total for all arrangements automatically finds the size of the second group, without needing to place the separated letters directly.

Combining constraints with repeated letters

A word can have both a togetherness requirement and repeated letters at the same time, and the two techniques combine directly. Glue the required letters into a block as before, but now check whether any of the glued letters are identical to each other or to letters outside the block, and divide by the appropriate factorial wherever a repeat occurs, both inside the block and among the remaining items.

A useful special case appears when the letters required to stay together are themselves identical, such as two copies of the same letter. In that case, the block has only one possible internal order, since swapping two identical letters inside the block does not create a new arrangement, so no extra factor is multiplied in for the block's internal order.

See the reasoning unfold

Pause after each step and predict what comes next.

Common mistakes to catch

• Forgetting to multiply by the block's internal arrangements after gluing two different required letters together.
• Multiplying by an extra $2!$ for a glued block whose two letters are identical, when that block actually has only one internal order.
• Trying to place separated letters directly instead of using the simpler total-minus-together approach, which often leads to missed or double-counted cases.

Try with support

Guided practice

Choose a strategy, take a first step, and use the hints when you need direction.

Put it into practice

Independent practice

Work on your own first. If you reveal a solution, work through it and mark it reviewed.

Bring it together

• A togetherness requirement is handled by gluing the required letters into one block, then arranging the block with the remaining items.
• A separation requirement is handled by subtracting the glued-together count from the total unrestricted count.
• When the letters glued into a block are identical to each other, the block needs no internal-order factor, unlike a block of different letters.