Arranging things in order — and order matters, so UPCAT and UPCTA are different.
How many arrangements
all five letters, every time
Five different letters filling five slots: 5! = 5 × 4 × 3 × 2 × 1 = 120. D is 6! and B squares instead. Every letter is used once and order matters, which is exactly a permutation of all five.
This one question decides which formula you use.
U, P, C, A, T — five different letters
use n!
use n!repeats!
Write the letters out and tick them off. UPCAT has five, all different, so nothing needs dividing away.
| Slot | Choices left | Why |
|---|---|---|
| slot 1 | 5 | any letter |
| slot 2 | 4 | one gone |
| slot 3 | 3 | two gone |
| slot 4 | 2 | three gone |
| slot 5 | 1 | the last one |
Each letter used up shrinks the next slot by one.
5 × 4 × 3 × 2 × 1
A letter cannot be used twice in one arrangement. That is the only difference from the counting principle you already know.
5 × 5 × 5 × 5 × 5 = 3125 lets a letter come back, so it counts “words” like UUUUU and PAPAT that reuse letters. Each slot must lose the letter already placed: 5 × 4 × 3 × 2 × 1 = 120.
5 × 4 × 3 × 2 × 1 = 5!
read “five factorial”
| n! | Means | Value |
|---|---|---|
| 3! | 3 × 2 × 1 | 6 |
| 4! | 4 × 3 × 2 × 1 | 24 |
| 5! | 5 × 4 × 3 × 2 × 1 | 120 |
Each one is the row above times the new number.
Factorials grow fast — 10! is already 3 628 800.
5! = 5 × 4 × 3 × 2 × 1
write the chain
= 20 × 3 × 2 × 1
5 × 4 = 20
= 60 × 2 × 1
20 × 3 = 60
= 120
60 × 2 = 120
Multiplying by 1 at the end changes nothing, so many people stop at × 2. The answer is the same.
Swapping two identical letters gives the same word.
L twice, E twice, V once
arrangements = 5!2! · 2! = 1204 = 30
they swap invisibly
same story again
UPCAT has no repeats, so its bottom would be 1! · 1! · … = 1 and the answer stays 120.
Only 3 of the 10 get places, so stop the chain after 3 slots.
10 × 9 × 8 = 720
three ordered slots
= 10!7! = 10P3
the 7! cancels the unused tail
Gold to Ana and silver to Ben is different from the reverse, so this is a permutation. If the three received identical prizes, it would be a combination and six times smaller.
the girls become one block: [GG] B B B B
5 things to arrange
arrange the 5 things: 5! = 120
the block moves as one
arrange inside the block: 2! = 2
the girls can swap
120 × 2 = 240
both stages
“Not together” is the complement: all 6! = 720 seatings minus the 240 with the girls together leaves 480.
120 treats the block as one fixed object. The people inside it can still change places, so multiply by 2!.
No two together. 4 boys and 3 girls stand in a row. In how many ways can they stand if no two girls are side by side?
boys first: 4! = 24
_ B _ B _ B _ B _
5 gaps for 3 girls: 5P3 = 5 × 4 × 3 = 60
at most one girl per gap
24 × 60 = 1440
both stages
What is different: with three or more to keep apart, “all minus together” no longer works in one step. Seat the others first, then drop the kept-apart people into the gaps.
Alternating. 3 boys and 3 girls sit in a row, boys and girls alternating.
patterns: BGBGBG or GBGBGB
2 ways to start
boys 3! = 6, girls 3! = 6
each group fills its own seats
2 × 6 × 6 = 72
all three stages
What is different: the pattern itself is a choice. With equal numbers either group can start; with 4 boys and 3 girls only BGBGBGB works, and the factor 2 disappears.
3! × 3! = 36 fixes a boy in the first seat. Starting with a girl gives 36 more arrangements.
n is the unknown. Find n if nP2 = 56.
n(n − 1) = 56
two slots: n, then n − 1
n² − n − 56 = 0 → (n − 8)(n + 7) = 0
factor
n = 8
n counts objects, so not −7
What is different: write nPr as its short chain, not with factorials, and solve. Quicker still: 56 is two consecutive whole numbers multiplied, 8 × 7.
Fixed ends. How many arrangements of PLANET begin and end with a vowel?
ends: 2 × 1
the vowels A and E, filled first
middle: 4! = 24
the 4 letters left, in 4 slots
2 × 24 = 48
both stages
What is different: some slots have their own short list. Fill those first, then arrange the rest freely. One fixed letter is the simplest case: PLANET starting with P leaves 5! = 120.
first digit: 5 choices
anything but 0
then 5 × 4 × 3
0 is allowed in the other places
5 × 5 × 4 × 3 = 300
Or remove the bad ones: all arrangements 6P4 = 360, minus those starting with 0, which number 5P3 = 60: 360 − 60 = 300 ✓.
Two conditions at once: split by the first digit. 3-digit even numbers greater than 400 from 1, 2, 3, 4, 5, 6, no repeats. The hundreds digit is 4, 5 or 6, and whether it is even changes the units count:
hundreds 4: units 2 or 6, tens 4 → 2 × 4 = 8
4 is used up
hundreds 6: units 2 or 4, tens 4 → 8
6 is used up
hundreds 5: units 2, 4 or 6, tens 4 → 12
an odd lead takes no even digit
8 + 8 + 12 = 28
separate cases add
Permutation items rarely ask for a plain n!. Expect officers chosen from a club (nPr), people or books kept together (a block), digits where 0 cannot lead, or codes with no repeated letter. Ask first whether order matters; if it does not, it is a combination. Ten-second check: a restriction must make the count smaller than the unrestricted one.
Adding one more letter multiplies the count by that number.
2 · 3 = 6, 6 · 4 = 24, 24 · 5 = 120
How many distinct arrangements of
Six letters would give 6! = 720 — but the three A's are interchangeable and so are the two N's. Divide by 3! and 2!: 720 ÷ (6 × 2) = 60. C forgets to divide, counting swaps of identical letters as different words.
Hint: two different letters repeat here, and one of them appears three times. Both go on the bottom of the fraction.
