PnC Ordered UnorderedPnC Ordered Unordereda+b+c=48 , natural number solutionfind ordered and unordered solutionssolution- a+b+c=48natural number of solutions means a>=1b>=1 c>=1so a+b+c=45so total number of solutions will be45+3-1c3-1=47c2this is total number of solutions, will contain some cases when a,b,c are distinct (( let those are x cases)), when 2 are same and one is different (( y)) and when all are same ((z)) here x+y+z will give us unordered ))((like we already done in questions like we want to make 4 letters word from ALLAHABAD"))case1- when all are differentform will be like abc so total way to arrange these are 3!=6 and let cases when all r different is x so 6x casescase2- when 2 are same and one is different aab form so ways to arrange is 3!/2!=3(0,0,45))(1,1,43))......(22,22,1) total cases 23 but one case will be here when (15,15,15) all are same we have to remove thatso cases =22*3case3- when all are same = (15,15,15) only one case and way to arrange that is 3!/3!=1and sum of these three cases will give us total cases47c2=6x+3*22+1so 47c2-67=6xso x=169so x+y+z=169+22+1=192 this is unordered positive

2nd method-

i already told in ordered and unordered cases=ORDERED - 3 * those occuring twice - those occurng once]/3! + those occuring twice + those occuring once

((47c2-3*22-1))/6 +22+1=192

question3-in how many ways 20 identical apples can be distributed among aashna , rishith,jaskrit such that aashna>rishitha>jaskrit> and all will get atleast one apple

Ans: a+b+c=20

atleast 1

so a+b+c=17so total cases 17+3-1c3-1=19c2

now this will include cases of all distinct , 2 same one distinct and all same

so 6x+3y+z=19c2

2 same and one different

(0,0,17)

(1,1,15))

(2,2,13)

(8,8,1)

so 9 cases

and 9and all same there will no case

so 6x=19c2-3*9-0

so 6x=19c2-27

6x are total cases when all are different but we want case when a>rso we wwant only xso x=19c2-27/6=19*9-27/6=9*18/6=24 ans

(now when we want case when 2 same and one different

then

a+a+c=17

so 2a+c=17so c=17-2a

so 9 cases will be there

and arrangement in3!/2!=3 ways

so 9*3=27 cases)this is shortcut to find when 2 numbers are same

Question:find number of solutions of a+b+c=30 where a, b , c are distinct and positive integers and aAns:positive

so a+b+c=27

so 27+3-1c3-1=29c2

so 6x+3y+z=29c2

a+a+c=27

so 2a+c=27so c=27-2a

so c=14 cases but this will also include case when all are same so 13 cases

and all same one case

3a=27so a=9

we want to find x only here bcz aso x=29c2-3y-z/(6)

x=29c2-3*13-1/6

29*14-40/6=61

and 61+13+1 will be unordered case

but we want only case when a>b>c

so we only want cases when are are different(a+b+c=27 hai i already gave 1,1,1 to a,b,,c)

question-find ordered number of positive solutions of abc=30method-1abc=54030=2^2*3^3*5^1a^x*b^y*c^z=2^2*3^3*5^1a will some number which will contain 2 ,3 and 5same for band same forand there will be some power of 2 ,3 and 5 in them whatever will be the power 0,1,2 or 3but sum of the power of 2 will be 2 , sum of the power of 3 will be 3 and sum of the power of 5 will be 1so x+y+z=2((where x,y,z could vary from 0 to2)so number of ways=2+3-1c3-1=4c2x+y+z=33+3-1c3-1=5c2x+y+z=1so 1+3-1c3-1=3c2so total number of positive solutions will be=4c2*5c2*3c2=180so ordered number of positive solutions will be 180now if we want total number of integral solutions then we will find negative solutions alsofor negative out of a,b,c ,2 numbers will be negativeso choose 2 numbers out of those 3 numbersso 3c2*((positive solutions))so 3c2*180so total ordered solutions =180+3c2*180

abc=300

abc=2^2*3*5^2

so x+y+z=2 so 4c2x+y+z=1 so 3c2x+y+z=2 so 4c2

so positive ordered

= 4c2*3c2*4c2

6*6*3=108 positive ordered

and total 108+3c2*108

Unordered:cases when all are different

cases when 2 same one different

cases when all are same

a*b*c=2^2*3*5^2

when 2 are same

a^2*c=2^2*3*5^2

(1)^2*(X)(2)^2*(X)(5)^2*(X)(10)^2*x

so 4 cases will be there and way to arrange them is 3!/2!=3 cases

and when all are same

a^3=2^2*3*5^2not possible

so 6x+y+z=108

so x=108-12/6

so x=16

and y=4

so unordered =x+y=20

question:If a*b*c*d=648, then how many ordered integral values of a,b,c,d are possible...

Ans:648=2^3*3^4

so x+y+z+k=3 so 6c3=20

and x+y+z+k=4 =7c3=35

so total positive =700

now cases can be

all positive=700

2 positive 2 negative =4c2*700

or all negative =700

so 700+700+6*700

4200+1400=5600

a^x*b^y*c^z*d^k=2^3*3^4

now a ,b,c,d all will contain 0,1,2,or 3 power of 2 and sum of all those will be equal to 3

so x+y+z+k=3

and same for 3

so x+y+z+k=4

question:In how many ways can 5^17 be written as a product of three positive numbers(( ye sayad ek din kisi ne poocha tha ))

Ans:a+b+c = 17

19C2 = 171

when two are same :2a+c = 17total 9 solutions

(171-3*9)/ 3! + 9 = 33

cases ki bhi need nahi hai

x+y+z=17

when 2 are equal then

2x+y=17

so 9 cases

or make cases

(1)^2*x

(5^2)*x

(5^4)*x

(5^6)*x

,..(5^16)*x

in how many ways can 3^11 can be expressed as a product of 3 numbersa) 16 b) 27 c) 82 d) 10

question:In how many ways 1001^2 can be written as product of 3 integerstotal ordered and unordered

Ans:a*b*c=7^2*11^2*13^2

x+y+z=2=4c2=6x+y+z=2=4c2=6

x+y+z=2=4c2=6

so 6^3+3*6^3

4*6^3=864

now unordered

2 same

a^2*b=1^2*x7^2*x11^2*x13^2*x(7*11)^2*x(7*13)^2*x(11*13)^2*x(7*11*13)^2

same 8 for negative so 16

so 864-16*3/6 +16question:in how many ways 1000 can b xpressed as the product of 3 integers?ordered and unordered

Ans:1000=2^3*5^3

so x+y+z=3

so 5c2=10

and x+y+z=3

so 10

so 10*10=100

and negative 3*100=300

so total =400

now unordered

same a^2*b=2^3*5^3

1^2*x

2^2*x

5^2*x

10^2*x

(-1)^2*x

(-2)^2*x

(-5)^2*x

(-10)^2*x

so total y=8

but one case will be all same here so y=7

and z=1

so 6x+3*7+1=400

so x=400-22/6

so unordered=x+y+z ... this is total number of ordered and unordered solutions