Overview
Probability is a comparison between favorable outcomes and the outcomes that remain possible. Most errors in counting questions arise before any fraction is simplified: the outcomes are counted in inconsistent ways, or a condition is applied only to the numerator. We will develop an outcome model for each problem, make the order convention explicit, and use a block argument when two objects must remain together.
Choose ordered or unordered outcomes consistently
A bag contains four blue and three gold tokens, and two are drawn without replacement. To find the probability of different colors, count unordered pairs of individual tokens. There are C(7,2) = 21 possible pairs, each equally likely. A mixed pair is determined by choosing one of four blue tokens and one of three gold tokens, giving 4·3 = 12 favorable pairs. The probability is 12/21 = 4/7.
You could also count draw orders, but then both numerator and denominator must be ordered. A blue-then-gold sequence and a gold-then-blue sequence are different ordered outcomes, although they form the same unordered pair. The incorrect answer 2/7 counts just one order while using the ordered draw denominator. The answer 24/49 allows the first token to return before the second draw. The phrase “without replacement” changes the available second draw, so it must appear in the model, not merely in your final explanation.
Apply a given condition to the denominator
Now the bag contains three red, two blue and one gold token, and three tokens are selected. You are told that exactly two selected tokens have the same color. Labeling individual tokens makes the basic three-token selections equally likely. There are twenty unrestricted selections, but the stated condition removes some of them, so twenty is no longer the denominator.
The eligible color patterns are RRB, RRG, BBR and BBG. There are six RRB selections because the red pair can be chosen three ways and the blue token two ways. RRG contributes three, BBR contributes three, and BBG contributes one. The restricted sample space therefore contains thirteen selections. Four of them contain gold: the three RRG selections and the one BBG selection. The conditional probability is 4/13. Color patterns themselves are not equally likely; a pattern with three underlying selections carries three times the weight of one with only one.
Keep required neighbors together with a block
Five different books are arranged in a row, and two specified books must be adjacent. Temporarily tie the specified pair into one block. You now arrange four objects: that block and the three remaining books. They have 4! = 24 orders. Inside the block, the two books may appear in either order, so each outer arrangement corresponds to two actual book arrangements. The total is 24·2 = 48.
The block method works because every allowed arrangement has one identifiable adjacent pair, and compressing that pair produces exactly one arrangement of the four larger objects. Expanding the block restores the two internal possibilities. This one-to-one description guards against overcounting. The answer 24 forgets internal order, while 120 counts all unrestricted arrangements and never imposes adjacency. It is useful to state what a single counted object represents before multiplying counts together; otherwise, a correct factorial can be attached to the wrong collection.
Audit the counting model
In the conditional token problem, 1/5 comes from four favorable selections divided by all twenty. Its numerator is sound, but its denominator includes outcomes ruled out by the information already known. Conversely, treating the four color patterns as four equally likely events ignores how many labeled selections each represents. Both errors are repaired by one habit: define an elementary outcome whose likelihood is uniform, then apply every condition to that collection.
Before solving a new counting problem, decide whether order matters, whether objects are distinct, whether replacement occurs, and whether the question conditions on information. After counting, inspect one outcome to ensure it was neither omitted nor counted more than once. For required adjacency, test the compression-and-expansion argument; for probability, check that favorable outcomes are a subset of the denominator. In practice, write a sentence naming that denominator before calculating. That sentence often determines the entire problem more reliably than starting with a formula.
Apply the method to a new case
Three distinct letters A, B and C are arranged in a random order, with all six orders equally likely. Given that A occurs before B, what is the probability that A and B are adjacent? Condition first: the permitted orders are ABC, ACB and CAB. In ABC and CAB the letters A and B are adjacent, so the conditional probability is 2/3. The denominator is three, because knowing their order removes BAC, BCA and CBA. A calculation that counts all four adjacent arrangements but divides by all six gets the unconditional probability; it happens to give the same fraction here, yet it has not used the condition and cannot justify the requested event. Check the reasoning rather than the coincidentally equal number by listing the conditional sample space. Each surviving order has equal probability because the original six outcomes were equally likely.