10.1 Fundamental Counting Principle & Factorials

What GMAC Tests

Official Guide Math Review 3.4 puts counting on the Quantitative Reasoning map. On the current GMAT (Focus Edition) that does not mean contest combinatorics. You will not prove binomial identities, count lattice paths, or run inclusion-exclusion on four overlapping sets. You will count outcomes of a short process: codes, lineups, integers with digit restrictions, or rearrangements of a word. The section is 21 Problem Solving questions in 45 minutes with no calculator, so GMAC writes items that reward a clean slot diagram rather than a formula dump.

The engine is the Fundamental Counting Principle (FCP). Factorials are the FCP applied to "fill every remaining seat." Identical letters are the FCP with an overcount correction. Almost every later permutation or combination formula is this same multiplication, rewritten.

The Fundamental Counting Principle

If a process is a sequence of choices, multiply the number of options at each step.

If a task can be done in m ways or in n other mutually exclusive ways, add.

Worked example. A consultant packs 4 jackets, 3 shirts, and 2 pairs of shoes. An outfit is one of each. Number of outfits = 4 × 3 × 2 = 24. Each jacket pairs with every shirt and every shoe, so the counts multiply.

Worked example. A 5-character password is one letter A–Z followed by four digits 0–9. Letters and digits may repeat. First slot 26 options, each digit slot 10 options: 26 × 10 × 10 × 10 × 10 = 260,000. Replacement is allowed, so the 10 does not drop.

Worked example that adds. A 2-character code is either two letters from {A, B, C} (repeats allowed) or two digits from {1, 2, 3, 4} (repeats allowed), never a mix. Letter codes: 3 × 3 = 9. Digit codes: 4 × 4 = 16. Total = 9 + 16 = 25. Multiplying 9 × 16 would count mixed codes the stem forbade.

"Independent slots" here means the menu at this step does not require a case split. It is not yet the probability word from 10.3. If choosing 0 in the units place changes how many nonzero choices remain for the leading digit, you still multiply inside each case, then add the cases.

The Slot Method

Draw a blank for each position. Fill the most restricted blank first so you do not paint yourself into a corner.

Worked example: 3-digit integers (the hundreds digit cannot be 0), digits may repeat.

• Hundreds: 9 choices (1–9)
• Tens: 10 choices (0–9)
• Units: 10 choices (0–9)

Total: 9 × 10 × 10 = 900.

If digits cannot repeat:

• Hundreds: 9 choices (1–9)
• Tens: 9 remaining (10 digits minus the one already used, and 0 is now available)
• Units: 8 remaining

Total: 9 × 9 × 8 = 648. The tens slot is 9, not 8, because 0 joins the menu after you used a nonzero hundreds digit. The trap product 9 × 8 × 7 = 504 would be correct only if the digits came from 1–9, with 0 banned from every slot.

Worked example: 4-digit integers with four distinct digits.

• Thousands: 9 (1–9)
• Hundreds: 9 (0–9 except the thousands digit)
• Tens: 8
• Units: 7

Total: 9 × 9 × 8 × 7 = 4,536. The trap is 10 × 9 × 8 × 7 = 5,040, which counts strings such as 0751 as four-digit integers. They are not. Another trap is 9 × 8 × 7 × 6 = 3,024, which forgets that 0 becomes legal after the leading digit is placed.

First-digit nonzero, last-digit even

When two positions have restrictions, fill the tighter one first, and split cases if the remaining count depends on which option you took.

Worked example: even 3-digit integers with distinct digits. Even units digits: 0, 2, 4, 6, 8.

Case A: units digit is 0 (1 way).

• Hundreds: 9 choices (1–9)
• Tens: 8 remaining
• Subtotal: 9 × 8 = 72

Case B: units digit is 2, 4, 6, or 8 (4 ways).

• Hundreds: 8 choices (1–9 except the units digit; 0 is still banned in front)
• Tens: 8 remaining (10 digits minus units minus hundreds, and 0 is available)
• Subtotal: 4 × 8 × 8 = 256

Total: 72 + 256 = 328.

A single product cannot encode both "units already used 0, so hundreds has all 9 nonzero digits" and "units used 2, so hundreds has only 8 nonzero digits." Cases exist to keep each product honest.

Shorter restriction example. Codes of length 3 from {A, E, I, O, B, C, D}, first character a vowel (A, E, I, O), no repeats.

• Slot 1: 4 vowels
• Slot 2: 6 remaining characters
• Slot 3: 5 remaining

Total: 4 × 6 × 5 = 120. Filling the vowel slot first is the whole method.

Factorials

n! = n × (n − 1) × ... × 2 × 1 for a positive integer n.

0! = 1 by definition. There is one way to arrange an empty list: do nothing. The empty product is 1. This is not a curiosity. It is why C(n, 0) = 1 and P(n, 0) = 1 — one way to choose nothing, one way to arrange nothing.

Memorize through 6! (and 7! = 5,040 if you can). Quantitative Reasoning has no calculator. GMAC knows 5! = 120 and 6! = 720; those values appear as choices and as intermediate products.

Worked example. Five distinct books on a shelf: 5 × 4 × 3 × 2 × 1 = 120 = 5!. The first position has 5 choices, the next 4, and so on. A factorial is the slot method when every object is used once.

Worked example. Four distinct awards given to four distinct employees, one award each: 4! = 24. If only three of four awards are given to four employees, that is no longer 4! as a raw "use everyone" count; it is the permutation P(4, 3) = 4 × 3 × 2 = 24, which 10.2 names. Notice P(4, 3) = 4! / 1! = 24, and P(4, 4) = 4! / 0! = 24. 0! in the denominator is what makes P(n, n) = n! work.

n! grows fast. You will not be asked for 10! as a decimal. You will cancel factors: 8! / 6! = 8 × 7 = 56. Cancel before multiplying. 7! / 5! = 7 × 6 = 42, not 7! divided by longhand 120 on the page.

Identical Items

If some of the n objects are indistinguishable, n! overcounts. Every shuffle among the identical copies looked different in n! and is the same real string.

Distinct arrangements = n! / (k1! × k2! × ...) where ki is the size of each identical group.

Worked example. BANANA has 6 letters: A appears 3 times, N appears 2 times, B appears once.

6! / (3! × 2!) = 720 / (6 × 2) = 60.

If all six letters were distinct you would have 720 strings. The three A's can be shuffled 3! = 6 ways inside any string without changing it, and the two N's can be shuffled 2 ways, so divide by 12.

Worked example. TEETH has 5 letters: T twice, E twice, H once.

5! / (2! × 2!) = 120 / 4 = 30.

Worked example. SUCCESS has 7 letters: S three times, C twice, U once, E once.

7! / (3! × 2!) = 5,040 / (6 × 2) = 420.

The same correction applies to identical objects that are not letters: 5 flags in a row of which 3 are red and 2 are blue, otherwise identical, is 5! / (3! × 2!) = 10. That 10 is also C(5, 3), which 10.2 makes official: choosing 3 positions out of 5 for the red flags is the same count.

If every object is distinct, the denominator is 1! × 1! × ... = 1, and you recover n!. If all n objects are identical, n! / n! = 1: there is only one string of n copies of the same letter.

Traps on Quant

• Multiplying two disjoint case counts (should add).
• Adding the options in two sequential slots (should multiply).
• Treating 0abc as a 4-digit integer: leading zeros are banned when the object is an integer with a stated number of digits.
• Forgetting that after a nonzero leading digit is used, 0 is available in later slots.
• Using n! when k letters repeat — divide by k!.
• Forgetting 0! = 1, then stalling at P(n, n) or C(n, 0).
• Computing 9 × 8 × 7 for a 3-digit integer from digits 0–9 with distinct digits, which bans 0 in every slot instead of only in front.