Permutations — nPr, arrangements with restrictions: Question bank

2.7.10 · D5 · Maths › Statistics & Probability — Intermediate › Permutations — nPr, arrangements with restrictions

Pehle — symbol ko samjho

Is page par sab kuch $n P_{r}$ use karta hai. Khud ko trap-test karne se pehle, ensure karo ki tum isse do tareekon se padh sako.

Block (glue) aur gap methods jo puri jagah use hoti hain wo visual bhi hain. Ye do pictures dhyan mein rakho:

True or false — justify karo

Niche har answer ek poora reason hai. Ek simple "true/false" ka matlab hai ki tumne trap samjha nahi.

A B

aur

B A

do alag permutations count hote hain.

True. Ek permutation ek ordered arrangement hai, isliye do items ko swap karna genuinely nayi line-up deta hai — same letters, alag order.

A B

aur

B A

do alag combinations count hote hain.

False. Ek combination ek set hota hai (order ignored);

{A, B}

aur

{B, A}

same selection hai, isliye ye ek baar count hote hain — dekho Combinations — nCr.

n P_{r}

hamesha

n C_{r}

ke barabar hota hai.

False.

n P_{r} = r! \cdot n C_{r}

(pehle

r

objects choose karo, phir unhe order karo), isliye ye tabhi barabar hain jab

r! = 1

, yaani

r = 0

r = 1

. Warna

n P_{r}

bada hota hai kyunki wo

r!

orderings bhi count karta hai.

n P_{n} = n!

ke liye

0! = 1

zaroori hai.

True.

n P_{n} = \frac{n !}{( n - n )!} = \frac{n !}{0 !}

, aur ye sirf isliye

n!

deta hai kyunki

0!

1

define kiya gaya hai — dekho Factorials and 0!.

n P_{0} = 0

kyunki tum kuch arrange nahi karte.

False.

n P_{0} = \frac{n !}{n !} = 1

: kuch arrange na karne ka exactly ek tarika hai (empty arrangement), jaise ek empty photo line-up ka ek hi tarika hota hai.

"Together" restriction ke liye, do objects ko glue karna direct final answer deta hai.

False. Pair ko ek block maanne ke baad tumhe block ke internal

2!

orders se bhi multiply karna hoga, kyunki do glued objects block ke andar swap ho sakte hain (

A B

B A

) (dekho glue figure upar).

"Must be apart" ke liye gaps method aur "total − together" method alag answers dete hain.

False. Unhe agree karna chahiye — ye same arrangements ko do tareekon se count karte hain. Agar farq hai, to miscounting hui hai (usually "together" mein internal

r!

bhool gaye).

k

objects ko ek row mein arrange karne se hamesha exactly

k

gaps bante hain nayi items slot karne ke liye.

False.

k + 1

gaps bante hain: har adjacent pair ke beech ek aur har end par ek, e.g. _C_D_E_ mein

k = 3

ke liye

4

gaps hain (dekho gap figure upar).

5 P_{5}

aur

5 P_{4}

barabar hain.

True. Falling-product form mein,

5 P_{4} = 5 \cdot 4 \cdot 3 \cdot 2

aur

5 P_{5} = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1

; extra last factor

n - r + 1 = 5 - 5 + 1 = 1

hai, jo

1

se multiply karta hai aur kuch nahi badalta. Dono

120

ke barabar hain.

Jab objects ki repetition allowed ho tab

n P_{r}

use kar sakte hain.

False.

n P_{r}

distinct, no-repeat picks assume karta hai — exactly isliye falling product

n, n - 1, n - 2, \dots

shrink karta hai. Agar repeats allowed hain to pool har step par

n

hi rehta hai, jo

n^{r}

deta hai — dekho Permutations with Repetition.

Why questions

Hum har position par choices ko multiply kyun karte hain, add kyun nahi?

Kyunki position 1 fill karna aur phir position 2 aur phir … dependent decisions ki ek chain hai; "and-then" decisions multiply hoti hain — ye hai Fundamental Counting Principle. Ek tree imagine karo:

n

first-slot branches mein se har ek

n - 1

second-slot branches ugata hai, isliye leaves multiply hoti hain. Adding "either/or" alternatives count karta hai, sequential steps nahi.

Sabse restricted position pehle kyun fill karein?

Agar tum pehle aasaan slots par freedom kharch karte ho, to tight slot mein zero valid choices bachi ho sakti hain. Even-number example mein, pehle hundreds aur tens choose karne par sirf odd digits units ke liye bach sakti hain. Constraint pehle lock karna (units = even) guarantee karta hai ki wo hamesha satisfy ho sake.

"Together" method objects ko ek block mein kyun glue karta hai?

Glue figure dekho: ek block jo adjacent rehna chahiye wo ordering ke liye ek single unit ki tarah behave karta hai; gluing "

k

objects with adjacency rule" ko "

k - 1

ordinary objects" mein badal deta hai. Phir hum block ke do faces ko correct karte hain internal

r!

se multiply karke.

"Apart" case mein aksar complement (total − together) kyun use hota hai?

"Never together" exactly "together" ka opposite hai. Har arrangement ya to together hai ya apart, beech mein kuch nahi, isliye together-count ko total se subtract karne par precisely apart-count milta hai — usually direct gap-count se aasaan, aur dono agree karne chahiye.

n P_{r}

ko falling product

n (n - 1) \dots (n - r + 1)

ki jagah

\frac{n !}{( n - r )!}

kyun likha jata hai?

Ye same number hain. Falling product ko

\frac{( n - r )!}{( n - r )!}

se multiply karne par wo poora factorial

n!

upar ban jaata hai jabki untouched tail

(n - r)!

niche hota hai aur us part ko cancel karta hai jo tumne kabhi use nahi kiya. Ratio sirf shrinking product ka ek compact naam hai jo tum actually compute karte ho.

Ek extra constraint usually count kyun reduce karta hai?

Ek restriction kisi step par valid choices remove karti hai (product mein ek chhota factor), aur chhote factor se multiply karne par chhota product milta hai. Ye arrangement count kabhi increase nahi kar sakta — best case (ek constraint jo kuch forbid nahi karti) mein wo use unchanged rakhta hai.

Permutations probability mein denominator kyun ban sakte hain?

Jab har arrangement equally likely ho, to arrangements ki total sankhya hi sample space ka size hoti hai, jo favourable arrangements ki count ke niche baithti hai — dekho Probability — Equally Likely Outcomes.

Edge cases

n P_{r}

kya hota hai jab

r > n

0

hai: falling product

n (n - 1) \dots

eventually factor

0

tak pahunch jaata hai (objects khatam ho jaate hain), isliye poora product zero hai — tum available distinct objects se zyada arrange nahi kar sakte.

n P_{1}

kya hai?

Exactly

n

— falling product mein ek hi factor

n

hai; ek object ko ek slot mein arrange karna bas

n

objects mein se kaunsa place karna hai ye choose karna hai, order ke baare mein sochna nahi.

0 P_{0}

kya hai?

1

, kyunki

\frac{0 !}{0 !} = \frac{1}{1} = 1

— single empty arrangement,

n P_{0} = 1

ke saath consistent.

Agar saare

n

objects identical hain, to distinct permutations kitni hain?

Sirf

1

: koi distinguishable objects nahi hone par, har "arrangement" same lagta hai. Falling product ke peeche distinctness assumption collapse ho gayi hai — dekho Permutations with Repetition.

Ek multiset ke liye — jaise

M I S S I S S I P P I

ke letters (11 letters, repeats ke saath) — kya

n P_{r}

sahi tool hai?

Nahi.

n P_{r}

assume karta hai ki saare objects distinct hain. Repeated objects ke saath pehle pretend karo ki wo distinct hain (

11!

) aur phir har repeated group ke over-count ko divide out karo:

\frac{11 !}{4 ! 4 ! 2 !}

(chaar

S

, chaar

I

, do

P

ke liye). Ye general rule hai jiska "identical pair" case sabse chhota instance hai — dekho Permutations with Repetition.

"At least one vowel at an end" ke liye, direct counting better hai ya complement?

Complement usually cleaner hota hai: total arrangements count karo, phir wo subtract karo jinmein koi bhi vowel kisi bhi end par nahi hai, kyunki "at least one" kai overlapping cases span karta hai jo directly easily double-count ho jaate hain.

Agar do required-together objects identical hain, to kya tum block ke andar phir bhi

2!

se multiply karte ho?

Nahi. Identical objects ko swap karne se koi nayi arrangement nahi banti, isliye internal factor

\frac{2 !}{2 !} = 1

hai;

2!

se multiply karna over-count karta — same multiset correction jaise

M I S S I S S I P P I

, bas size

2

ke group ke saath.

Flashcards

n P_{r}

ki do readings kya hain?

Falling product

n (n - 1) \dots (n - r + 1)

(

r

shrinking factors) aur ratio

\frac{n !}{( n - r )!}

— same number.

n P_{r} = r! \cdot n C_{r}

kyun hai?

Ek ordered arrangement choose karke

r

objects (

n C_{r}

) aur phir unhe order karke (

r!

) banao; choose-aur-phir-order multiply karta hai.

n P_{r}

bada hai ya

n C_{r}

se chhota (for

r \geq 2

Bada,

r!

ke factor se, kyunki permutations wo orderings bhi count karte hain jo combinations ignore karti hain.

n P_{r}

kya hota hai jab

r > n

0

— falling product ek zero factor tak pahunch jaata hai; jo distinct objects exist karte hain unse zyada arrange nahi kar sakte.

k

arranged objects se kitne gaps bante hain?

k + 1

(har pair ke beech aur dono ends par).

Multiset ke permutations (repeated objects)?

\frac{n !}{n _{1} ! n _{2} ! \dots}

— full factorial ko har repeated group ke size ke factorial se divide karo.

Jab glued pair identical ho to kya internal

2!

se multiply karte hain?

Nahi — identical objects koi nayi order nahi dete, isliye factor

1

hai.

Connections

Permutations — nPr, arrangements with restrictions — parent topic jiske ye traps target hain.
Fundamental Counting Principle — kyun hum add nahi multiply karte.
Combinations — nCr — "order ignored" contrast; $n P_{r} = r! \cdot n C_{r}$ bridge ka source.
Factorials and 0! — $0! = 1$ edge cases.
Circular Permutations — round-table trap.
Permutations with Repetition — multiset aur identical-object traps.
Probability — Equally Likely Outcomes — jahan ye counts denominators ban jaate hain.

Question bank — Permutations — nPr, arrangements with restrictions

Pehle — symbol ko samjho

True or false — justify karo

Error dhundho

Why questions

Edge cases

Flashcards

Connections