Permutations — nPr, arrangements with restrictions: Worked examples

2.7.10 · D3 · Maths › Statistics & Probability — Intermediate › Permutations — nPr, arrangements with restrictions

Shuru karne se pehle, ek reminder us EK tool ka jo sab kuch underlie karta hai: multiplication principle from Fundamental Counting Principle — "agar pehla choice $a$ tareekon se ho sakta hai aur agla choice $b$ tareekon se, toh dono mil ke $a \times b$ tareekon se ho sakte hain." Neeche har count bas yahi hai, slot by slot apply kiya hua.

The scenario matrix

Har permutation problem ko is table ke exactly ek cell mein girta hua socho. Sabse right column mein us worked example ka naam hai (E1–E9) jo wahan aata hai.

#	Case class	Tricky kyun hai	Covered by
A	Plain $n P_{r}$	bas $r$ slots fill karo	E1
B	Degenerate: $r = 0$ ya $r = n$	empty arrangement / saare objects; $0! = 1$ chahiye	E2
C	Degenerate: $r > n$	impossible cheez maangna → count hai $0$	E2
D	Restricted slot (pehle forced fill karo)	ek position sirf kuch objects accept karti hai	E3
E	Together (glue block)	internal $r!$ yaad rakhna zaroori	E4
F	Apart (gaps method)	gaps ko $k + 1$ count karo	E5
G	Fixed ends / dono ends forced	ek saath do constraints	E6
H	"At least" / forbidden (complement)	total − bad count karo	E7
I	Real-world word problem	words → slots mein translate karo	E8
J	Exam twist (mixed patterns + parity)	do restrictions ek saath interact kar rahi hain	E9

Har cell A–J neeche solve hoti hai. Agar tum sab nau kar sako, toh tumne literally is topic ki har shape dekh li hai.

E1 — Cell A · Plain $n P_{r}$

Worked example Statement

7 distinct letters ${P, Q, R, S, T, U, V}$ mein se, kitne ordered 4-letter strings (bina repeat ke) banaye ja sakte hain?

Forecast: guess karo ki answer $7^{4} = 2401$ se bada hoga ya chhota, aur kyun.

Step 1. Hum 4 positions fill karte hain, left se right, 7 letters mein se. Yeh step kyun? Order matter karta hai (string ordered hoti hai), no repeats → yeh permutation hai, combination nahi (Combinations — nCr order ignore karta).

Step 2. Har slot ke liye choices count karo: $7, 6, 5, 4$ . Yeh step kyun? Har rakha gaya letter use ho jaata hai, isliye pool har baar ek se ghatta hai — multiplication principle with dependent choices.

Step 3. Multiply karo: $7 P_{4} = 7 \cdot 6 \cdot 5 \cdot 4 = 840 = \frac{7 !}{3 !} .$ Yeh step kyun? Multiply karna counting principle hai; $\frac{7 !}{3 !}$ form same product ko repackage karta hai ( $3! = 3 \cdot 2 \cdot 1$ jo hum "tak nahi pahunche" woh cancel ho jaata hai).

Verify: $840 < 2401$ ✓ — $7^{4}$ se kam kyunki repeats banned hain, toh baad ke slots mein fewer options hain. Aur $\frac{7 !}{3 !} = \frac{5040}{6} = 840$ ✓.

E2 — Cells B & C · The degenerate inputs

Yeh woh cases hain jo students skip karte hain aur phir aasaan marks kho dete hain. Hum teeno ek saath lete hain.

Worked example Statement

Evaluate karo (a) $6 P_{0}$ , (b) $6 P_{6}$ , (c) $4 P_{7}$ .

Forecast: inme se ek $1$ hai, ek $720$ hai, ek $0$ hai. Padhne se pehle match karo.

Step 1 — (a) $6 P_{0}$ . 6 objects mein se zero slots fill karo. Yeh step kyun? Kuch bhi arrange na karne ka exactly ek tarika hai — empty arrangement. Formally $6 P_{0} = \frac{6 !}{6 !} = 1$ (isme $0! = 1$ use hota hai from Factorials and 0!). $6 P_{0} = 1.$

Step 2 — (b) $6 P_{6}$ . Saare 6 objects arrange karo. Yeh step kyun? Slots ko $6, 5, 4, 3, 2, 1$ choices milti hain, yani $6!$ . Formula form mein $\frac{6 !}{( 6 - 6 )!} = \frac{6 !}{0 !} = 6!$ . $6 P_{6} = 6! = 720.$

Step 3 — (c) $4 P_{7}$ . Sirf 4 mein se 7 chosen objects arrange karo. Yeh step kyun? Pool khatam ho jaayega: slots dete hain $4, 3, 2, 1, 0, \dots$ aur woh $0$ product ko khatam kar deta hai. Koi valid arrangement nahi hai. $4 P_{7} = 0.$

Verify: formula $\frac{4 !}{( 4 - 7 )!} = \frac{4 !}{( - 3 )!}$ negative number ke factorial ke liye undefined hai — common-sense answer se milta hai "impossible = $0$ arrangements." ✓

E3 — Cell D · Restricted slot, pehle forced position fill karo

Worked example Statement

${A, B, C, D, E}$ ko is tarah arrange karo ki arrangement consonant pe khatam ho.

Forecast: saari $5! = 120$ arrangements ke aadhe se zyada ya kam? (Hint: 5 mein se 3 letters consonants hain.)

Step 1. Pehle last slot handle karo: consonants hain ${B, C, D}$ → $3$ choices. Yeh step kyun? Constraint sirf last position pe hai. Sabse tight slot pehle lock karna ensure karta hai ki hum use kabhi unsatisfiable na chhodein (Pattern 1 / mnemonic "F for Forced first").

Step 2. Baaki 4 letters baaki 4 slots freely fill karte hain: $4! = 24$ . Yeh step kyun? Ek baar last letter choose ho gayi, baaki positions mein koi restriction nahi — jo bacha uska plain permutation.

Step 3. Multiply karo: $3 \times 24 = 72.$

Verify: symmetry se, consonant pe khatam hone ka fraction $= \frac{3}{5}$ , toh expected count $= \frac{3}{5} \times 120 = 72$ ✓. Aur $72 > 60$ , yani aadhe se zyada, jo "5 mein se 3 letters" se match karta hai. ✓

E4 — Cell E · Together (block glue karo)

Worked example Statement

Chhe log ${1, 2, 3, 4, 5, 6}$ row mein baithte hain. Kitni arrangements mein persons $1$ aur $2$ ek doosre ke saath hain?

Forecast: saari $6! = 720$ seatings mein se kitne fraction mein woh adjacent end up honge?

Step 1. $1$ aur $2$ ko ek block $[12]$ mein glue karo. Ab hamare paas 5 items hain: $[12], 3, 4, 5, 6$ . Yeh step kyun? Adjacency matlab woh ek unit ki tarah move karte hain — unke beech ka gap zero hai. Unhe ek object treat karna "adjacent" ko count mein automatically bake kar deta hai (Pattern 2). Figure mein teal mein outlined block dekho.

Step 2. Yeh 5 items arrange karo: $5! = 120$ . Yeh step kyun? Paanch distinct items ek row mein → plain permutation.

Step 3. Block ke andar, $1$ aur $2$ swap kar sakte hain: $2! = 2$ (orders $12$ aur $21$ ). Yeh step kyun? "Adjacent" yeh nahi kehta ki kaun left pe hai — dono internal orders valid arrangements hain. Figure dono internal orders side by side dikhata hai.

Step 4. Multiply karo: $5! \times 2! = 120 \times 2 = 240.$

Verify: fraction adjacent $= \frac{240}{720} = \frac{1}{3}$ — sensible hai: person 1 ke paas 6 seats hain; unki seat given, person 2 ke adjacent seats baaki 5 mein se 2 hain ends chhod ke... ek jaldi sanity check: $\frac{1}{3} \times 720 = 240$ ✓.

E5 — Cell F · Apart (gaps method)

Worked example Statement

${A, B, C, D, E}$ ko row mein arrange karo taaki $A$ , $B$ , aur $C$ mein se koi bhi do adjacent na hon (teeno mutually separated).

Forecast: kya teen items ko gaps mein daalne ke baad enough room bachega? Pehle apne dimaag mein gaps count karo.

Step 1. Pehle unrestricted letters $D, E$ arrange karo: $2! = 2$ . Yeh step kyun? Gaps method "must-be-apart" objects ko doosron ke banaye gaps mein rakhta hai. Isliye pehle free objects lay down karte hain (Pattern 3).

Step 2. Do objects $2 + 1 = 3$ gaps banate hain: $_D_E_$ (figure mein plum slots ki tarah dikhaaye gaye). Yeh step kyun? $k$ arranged objects se Gaps $= k + 1$ — dono ends including. $A, B, C$ mein se har ek ko ek alag gap mein daalna guarantee karta hai ki koi touch na kare.

Step 3. Hume 3 items 3 gaps mein rakhne hain, ek per gap: woh hai $3 P_{3} = 3! = 6$ . Yeh step kyun? Exactly 3 gaps aur 3 items, sab distinct, ek per gap → ordered placement = permutation.

Step 4. Multiply karo: $2! \times 3! = 2 \times 6 = 12.$

Verify: kyunki 3 items exactly 3 slots mein jaate hain toh har slot filled hoga, pattern hamesha $D ? E ?$ ... ek skeleton list karte hain: gaps $g_{1} D g_{2} E g_{3}$ , $A, B, C$ mein se ek har gap mein → $3! = 6$ fillings $\times 2$ orders of $D, E = 12$ ✓. (Note: agar hum sirf $A, B$ ko apart chahte toh zyada room hoti; yahan tight count $12$ sahi hai kyunki gaps exactly items se match karti hain.)

E6 — Cell G · Dono ends fixed (do constraints)

Worked example Statement

5 letters ${A, B, C, D, E}$ ko arrange karo taaki string vowel se start ho aur consonant pe khatam ho.

Forecast: ek saath do constraints — kya woh cleanly multiply hote hain, ya interfere karte hain?

Step 1. Pehla slot (vowel): ${A, E}$ → $2$ choices. Yeh step kyun? Do forced positions hain; constraints ek ek karke handle karo, sabse tight pehle. Dono ends equally tight hain, toh kisi se bhi shuru karo.

Step 2. Last slot (consonant): ${B, C, D}$ → lekin kya ek letter ab use ho gayi? Nahi — vowels aur consonants disjoint sets hain, isliye saare 3 consonants abhi bhi available hain → $3$ choices. Yeh step kyun? Humein check karna hai ki Step 1 ne Step 2 se koi option chheena ya nahi. Yahan nahi chheena (vowel kabhi consonant nahi hota), isliye counts independent rehte hain.

Step 3. Middle 3 slots: 5 mein se 3 letters bache → $3! = 6$ . Yeh step kyun? Dono ends fix karne ke baad, do letters use ho gayi hain; baaki 3 middle ko freely fill karte hain.

Step 4. Multiply karo: $2 \times 3 \times 3! = 2 \times 3 \times 6 = 36.$

Verify: total arrangements $= 120$ ; fraction $= \frac{36}{120} = \frac{3}{10}$ . Independent-fraction cross-check: P(start vowel) $= \frac{2}{5}$ , P(end consonant | start vowel) $= \frac{3}{4}$ , product $= \frac{6}{20} = \frac{3}{10}$ ✓.

E7 — Cell H · "At least" via complement

Worked example Statement

${A, B, C, D, E}$ ki kitni arrangements mein $A$ aur $B$ saath mein end up hon ya $A$ start pe ho (yani kam se kam ek condition hogi)?

Forecast: directly overlaps count karna messy hai — kya complement ya inclusion–exclusion cleaner hoga? Hum yahan inclusion–exclusion use karte hain.

Step 1. " $A, B$ together" count karo: glue block → $4! \times 2! = 48$ (E4 ki method jaisi, lekin 5 letters). Yeh step kyun? Yeh event $T$ hai. Hum glue method par already trust karte hain.

Step 2. " $A$ at start" count karo: $A$ pehle fix karo, baaki 4 arrange karo → $4! = 24$ . Yeh step kyun? Yeh event $S$ hai; ek forced position.

Step 3. Overlap count karo " $A, B$ together AND $A$ at start": $A$ position 1 pe hai, toh $B$ position 2 pe hona chahiye, phir $C, D, E$ free → $3! = 6$ . Yeh step kyun? Jab do events saath ho sakti hain, $∣ S \cup T ∣ = ∣ S ∣ + ∣ T ∣ - ∣ S \cap T ∣$ — warna overlap double-count hoga.

Step 4. Combine karo: $48 + 24 - 6 = 66.$

Verify: complement check — arrangements jisme koi bhi condition nahi $= 120 - 66 = 54$ . Direct count of "neither": $A$ not at start (positions 2–5, 4 choices) aur $A, B$ not adjacent. Sanity: total not-adjacent $= 120 - 48 = 72$ ; unme se jo $A$ start pe hain $= 24 - 6 = 18$ ; toh not-adjacent AND not-start $= 72 - 18 = 54$ ✓. Dono routes $66$ dete hain.

E8 — Cell I · Real-world word problem

Worked example Statement

Ek photographer 4 medals — gold, silver, bronze, aur ek special "wildcard" — ko shelf pe line up karta hai, lekin gold aur silver medals ek doosre ke saath nahi ho sakte (woh scratch ho jaayenge). Kitne valid displays hain?

Forecast: letters mein translate karo. Kaun sa method — glue, gaps, ya complement? Compute karne se pehle choose karo.

Step 1. Rename karo: $G, S, B, W$ ; $G, S$ adjacent forbidden hai. Yeh step kyun? Word problems bas permutations hain costume pehne hue; story ko strip karo — distinct objects + ek constraint.

Step 2. Total arrangements: $4! = 24$ . Yeh step kyun? Chaar distinct medals, sab placed → plain permutation.

Step 3. " $G, S$ together" count karo (subtract karne ke liye): glue → $3! \times 2! = 6 \times 2 = 12$ . Yeh step kyun? Complement yahan sabse aasaan hai (ek forbidden pattern). $G, S$ ko $B, W$ mein ek block mein glue karo = 3 items, times internal $2!$ .

Step 4. Valid $=$ total $-$ together: $24 - 12 = 12.$

Verify: gaps cross-check — pehle $B, W$ arrange karo ( $2! = 2$ ), $3$ gaps milte hain, $G, S$ ko alag gaps mein rakho $= 3 P_{2} = 6$ ; total $2 \times 6 = 12$ ✓. Units sahi lagte hain: displays ka integer count. ✓

E9 — Cell J · Exam twist (mixed patterns + parity)

Worked example Statement

Digits ${1, 2, 3, 4, 5, 6}$ use karke bina repeats ke, kitne 4-digit even numbers greater than 3000 ban sakte hain?

Forecast: do constraints interact kar rahi hain — leading digit (">3000" ke liye) aur units digit ("even" ke liye) dono same small even digits ke liye compete karte hain. Guess karo ki kya tum naively multiply kar sakte ho (nahi kar sakte).

Step 1 — ek smart variable pe split karo. Clash yeh hai: even units digit aur leading digit $\geq 3$ dono digit $4$ ya $6$ chahte hain. Isliye hum units digit pe case-split karte hain. Yeh step kyun? Jab do restrictions ek hi objects ke liye ladte hain, shared resource pe casework counts ko clean rakhta hai.

Step 2 — Case (a): units digit $= 2$ . Toh thousands digit $\in {3, 4, 5, 6}$ hona chahiye ( $> 3000$ ke liye, aur $\neq = 2$ ) → $4$ choices. Middle do digits baaki $4$ digits mein se → $4 \times 3 = 12$ . Yeh step kyun? Units fixed hai (2 $\geq 3$ set mein nahi, toh thousands se clash nahi); thousands count karo, phir do middle slots. $1 \times 4 \times 4 \times 3 = 48.$ Ruko — slots ko (thousands)(mid)(mid)(units) order karo: $4$ (thousands) $\times 4 \times 3$ (middles) $= 48$ .

Step 3 — Case (b): units digit $\in {4, 6}$ (2 choices). Ab units digit un $\geq 3$ digits mein se ek hai, isliye thousands ke saath compete karta hai. Thousands $\in {3, 4, 5, 6}$ mein se hona chahiye minus woh digit jo already units ke roop mein use ho gayi → $4 - 1 = 3$ choices. Middle do baaki $4$ digits mein se → $4 \times 3 = 12$ . Yeh step kyun? Yahan pehle ki choice ne actually thousands se ek option haata diya (E6 se alag) — isliye 1 subtract karo. Yahi woh trap hai jiske around twist build hua hai. $2 \times 3 \times 4 \times 3 = 72.$

Step 4 — Disjoint cases add karo: $48 + 72 = 120.$ Yeh step kyun? Dono cases (units $= 2$ vs units $\in {4, 6}$ ) kabhi overlap nahi karte, isliye hum add karte hain (Pattern 5 ka sibling — partition, phir sum).

Verify: sanity — total even 4-digit numbers (bina ">3000" rule ke) $= 3 (units even) \times 5 \times 4 \times 3 = 180$ ; hamara answer $120 < 180$ expected hai kyunki ">3000" kuch haata deta hai. Removed ones ka direct recount (thousands $\in {1, 2}$ , even): units even & thousands $\in {1, 2}$ ... $60$ deta hai, aur $180 - 60 = 120$ ✓.

Recall Main kis cell mein hoon? (quick triage)

Kya $r > n$ hai? ::: Answer hai $0$ (E2, cell C). Kya $r = 0$ ya $r = n$ hai? ::: Answer hai $1$ ya $n!$ — $0! = 1$ yaad rakho (E2, cells B). Kya ek position sirf kuch objects accept karti hai? ::: Pehle woh slot fill karo (E3, cell D). Objects adjacent hone chahiye? ::: Block mein glue karo, times internal $r!$ (E4, cell E). Objects sab separated hone chahiye? ::: Gaps method, $k + 1$ gaps (E5, cell F). Do forced ends? ::: Dono fill karo, check karo ki kya woh objects share karte hain (E6/E9, cells G/J). "At least" / "or" / "not"? ::: Complement ya inclusion–exclusion (E7/E8, cells H/I). Do restrictions same digits ke liye ladhti hain? ::: Shared resource pe case-split karo, phir add karo (E9, cell J).

Connections

Fundamental Counting Principle — yahan har multiply-the-choices step yahi rule hai.
Combinations — nCr — jab order matter karna band ho jaaye tab use karo (E1 ka contrast).
Factorials and 0! — degenerate cells B ko kaam karta hai ( $0! = 1$ ).
Circular Permutations — upar ke row problems ring mein $(n - 1)!$ ban jaate hain.
Permutations with Repetition — "no repeats" hata do aur counts change ho jaate hain.
Probability — Equally Likely Outcomes — hamare "Verify" steps mein jo fractions hain woh exactly yahi probabilities hain.