Try each one on paper first, then check a single step. That is worth far more than reading a finished solution.
Example 1 — arrangements with repeated letters
How many distinct arrangements are there of the letters of LETTER?
- 1There are 6 letters, so 6!=720 if all were distinct.
- 2But E appears twice and T appears twice.L, E, T, T, E, R.
- 3Swapping the two E's produces no new word, and neither does swapping the T's.So 6! counts each arrangement several times.
- 4Divide by 2! for the E's and 2! for the T's.
- 52×2720=180
Example 2 — an arrangement with a restriction
Five people sit in a row. In how many ways can they be arranged if two particular people must sit together?
- 1Treat the two people as a single block.The standard technique for 'together' questions.
- 2There are now 4 items to arrange: the block plus three individuals.
- 34!=24 arrangements of those items.
- 4Within the block the two can sit in 2!=2 orders.Easy to forget.
- 524×2=48And 'not together' would be 120−48=72.
Example 3 — probability without replacement
A bag holds 3 red and 5 blue balls. Two are drawn without replacement. Find the probability that one is red and one is blue.
- 1There are two orders: red then blue, or blue then red.Both must be counted.
- 2Red then blue: 83×75=5615After a red is removed, 7 balls remain, 5 of them blue.
- 3Blue then red: 85×73=5615Equal, as it happens.
- 4Add: 5630The two routes are mutually exclusive, so their probabilities add.
- 5=2815Check: P(2 red)+P(2 blue)+P(one of each)=283+2810+2815=1 ✓