Exercises — Permutations — nPr, arrangements with restrictions

HUM KYA karte hain: $3$ slots fill karte hain $7$ distinct objects mein se. YEH numbers kyun: slot 1 mein saare $7$ available hain; ek use hone ke baad, slot 2 mein $6$ hain; phir slot 3 mein $5$. $7P_3=7\times 6\times 5=210$ Compact form se check karo: $\frac{7!}{(7-3)!}=\frac{7!}{4!}=\frac{5040}{24}=210.$ ✓

Sab kuch arrange karna matlab $r=n$, toh hum chahte hain $6P_6=\frac{6!}{0!}$. $0!=1$ kyun: kuch bhi nahi arrange karne ka exactly ek hi tarika hai, isliye $0!$ se divide karne se kuch nahi badalta (dekho Factorials and 0!). $6P_6=6!=720$

Test yeh hai: kya donon chosen logon ko swap karne se alag outcome milta hai? Haan — (Ana President, Bo Secretary) $\ne$ (Bo President, Ana Secretary). Order matters $\Rightarrow$ permutation hai, Combinations — nCr nahi. $10P_2=10\times 9=90$

KYA restrict karta hai: "odd" sirf units digit ko touch karta hai, toh pehle woh slot lock karo (tightest constraint pehle).

• Units: odd hona chahiye $\Rightarrow \{1,3,5\}\Rightarrow 3$ choices.
• Thousands: bache hue $5$ mein se koi bhi $\Rightarrow 5$.
• Hundreds: $4$ remaining.
• Tens: $3$ remaining. $3\times 5\times 4\times 3=180$

$3$ doston ko ek block $[FFF]$ mein glue karo. Ab hum $4$ items arrange karte hain: block plus baaki $3$ log.

• $4$ items arrange karo: $4!=24$.
• Block ke andar $3$ dost apas mein reorder kar sakte hain: $3!=6$. $4!\times 3!=24\times 6=144$

Gaps method. Pehle baaki $3$ letters $R,S,T$ arrange karo: woh $3!=6$ hai. Figure dekho: teen placed letters $\underline{}R\underline{}S\underline{}T\underline{}$ create karte hain — matlab $3+1=4$ gaps.

Complement kyun: "at least one even" kai cases span karta hai (ek even, do even, teen even). Iska opposite — "koi bhi even digit nahi" — ek single clean case hai.

• Total $3$-digit numbers: $7P_3=7\times 6\times 5=210$.
• All-odd numbers sirf $\{1,3,5,7\}$ (woh $4$ odds) use karte hain: $4P_3=4\times 3\times 2=24$. $\text{at least one even}=210-24=186$

Pehle do forced slots lock karo, phir middle fill karo.

• Pehla slot (vowel): $3$ choices.
• Aakhri slot (consonant): $3$ choices.
• Beech ke $4$ slots: bache hue $4$ letters kisi bhi order mein $\Rightarrow 4!=24$. $3\times 3\times 4!=9\times 24=216$

Hidden restriction: leading digit $0$ nahi ho sakta, toh pehle thousands slot lock karo.

• Thousands: $\{1,2,3,4\}$ mein se koi bhi $\Rightarrow 4$ choices ($0$ forbidden hai yahan).
• Ab $0$ baaki ke liye wapas available hai. Remaining slots $4$ leftover digits se draw karte hain:
• Hundreds: $4$ choices.
• Tens: $3$.
• Units: $2$. $4\times 4\times 3\times 2=96$

Glue then subtract. $[AB]$ ko ek block treat karo; block $4!\times 2!=48$ total adjacent arrangements deta hai (parent E3). Ab woh arrangements nikalo jahan $A$ end par baithta hai. Block ke do orderings dekho:

• Block $=AB$: $A$ end par tabhi hota hai jab block sabse leftmost position par ho (toh $A$ position 1 par hai). Woh $1$ block-placement $\times 3!$ baaki ke liye $=6$ hai.
• Block $=BA$: $A$ end par tabhi hota hai jab block sabse rightmost position par ho (toh $A$ position 5 par hai). Phir se $3!=6$. Forbidden ($A$ end par, adjacent) $=6+6=12$. $48-12=36$

Gaps method, order mein karo.

• Pehle $4$ boys arrange karo: $4!=24$.
• Chaar boys $4+1=5$ gaps create karte hain: $\underline{}B\underline{}B\underline{}B\underline{}B\underline{}$.

  

• $3$ girls ko $3$ alag gaps mein place karo (alag gaps $\Rightarrow$ kabhi adjacent nahi), order matters: $5P_3=5\times 4\times 3=60$. $4!\times 5P_3=24\times 60=1440$

Complement zyada clean hai: "saath nahi" $=$ total $-$ "teeno saath".

• Total: $6!=720$.
• Teeno saath (glue $[ABC]$): $4!\times 3!=24\times 6=144$. $720-144=576$

Do restrictions collide karte hain: units even hona chahiye $\{0,2,4\}$, aur thousands $\ne 0$. Jab units $0$ leta hai, toh thousands slot kya use kar sakta hai woh badal jaata hai — isliye split karo units digit $0$ hai ya nahi us par.

Case A — units $=0$:

• Units: $1$ way ($0$).
• Thousands: bache hue $\{1,2,3,4,5\}$ mein se koi bhi $\Rightarrow 5$ (koi zero conflict nahi, $0$ already use ho gaya).
• Hundreds: $4$ remaining.
• Tens: $3$ remaining.
• Subtotal $=1\times 5\times 4\times 3=60$.

Case B — units $\in \{2,4\}$:

• Units: $2$ ways.
• Thousands: $0$ nahi ho sakta aur units digit ke barabar nahi ho sakta. $6$ digits mein se $0$ aur use kiya gaya even hatao $\Rightarrow 4$ choices.
• Hundreds: $4$ remaining (ab $0$ yahan allowed hai).
• Tens: $3$ remaining.
• Subtotal $=2\times 4\times 4\times 3=96$.

$60+96=156$

Do glue blocks.

• $3$ Maths ko block $M$ mein, $2$ Physics ko block $P$ mein glue karo. Arrange karne ke items: $M,P,C_1,C_2$ $\Rightarrow 4$ items $\Rightarrow 4!=24$.
• $M$ ke andar: $3!=6$. $P$ ke andar: $2!=2$. $4!\times 3!\times 2!=24\times 6\times 2=288$

Yeh seedha Probability — Equally Likely Outcomes se link karta hai: saare $6!$ orderings equally likely hain, toh probability $=\frac{\text{favourable}}{\text{total}}$ hai.

• Total: $6!=720$.
• Favourable: dono ends vowels hain. Exactly $2$ vowels $\{A,E\}$ aur $2$ end-slots hain. Unhe place karo: $2P_2=2!=2$. Beech ke $4$ letters: $4!=24$. Favourable $=2\times 24=48$. $P=\frac{48}{720}=\frac{1}{15}$

Yeh sab same move hai: khaali slots line up karo, aur har slot par poochho "abhi kitne allowed hain?" phir multiply karo. Restrictions sirf un counts mein se ek ko change karti hain. Jab do restrictions ek hi slot par push karti hain (even aur no leading zero), cases mein split karo taaki har case ke counts clean hon.

L1: $210$, $720$, $90$. L2: $180$, $144$, $72$. L3: $186$, $216$, $96$. L4: $36$, $1440$, $576$. L5: $156$, $288$, $\frac{1}{15}$.