Permutation & Combination

In how many ways we arrange 6 boys and 4 girls
a) such that all girls are together
b) all boys are together
c) all boys are together , all girls together
d) no girls are together

a) all girls are together so make a group of them and take it as a one unit
so they could be arranged in 4! ways so now we have BBBBBB(G) so total 7 persons , number of ways to arrange those 7!
so total ways =7! * 4

b) all boys are together so make it as one entity so they could be arranged in 6! ways
so total number of ways = 6! * 5!

c) all boys are together and all girls are together
so BBBBBBGGGG
6!*4!
and GGGGBBBBBB
so 4 * 6!
so total ways = 2 * 4! * 6!

d) no two girls are together
so _ B _ B _ B _ B _ B _ B _
first arrange 6 boys so ways 6!
now 6 boys will leave 7 spaces
now select 4 spaces and arrange 4 girls there
so 7c4*4!
so total ways = 6! * 7c4 * 4!

Five boys and five girls form a line with the boys and girls alternating . The number of ways of making the line

GBGBGBGBGB
so 5!*5!
now BGBGBGBGBG
so 5!*5!
so total ways = 5! * 5! + 5! * 5! = 2 * (5!)^2

NUMBER OF ARRANGEMENTS that can be made with the letters of the word " MATHEMATICS " in which all vowels come together

there are 4 vowels A,E,A,I , number of ways to arrange them = 4!/2!
now considering four vowels as one letter ( bcz we want them together ) we have 8 letters
M,T,H,M,T,C,S and one letter combining the vowels
so number of ways to arrange them = 8!/(2!)*2! ( bcz of 2 M and 2 T )
so total ways =8!/2!*2! * 4!/2!

Concept : HOW TO FIND RANK OF A WORD

Find the rank of the word " SACHIN" or SACHIN appears in at which number in dictionary

IN DICTIONARY The words at each stage are arranged in alphabetic order .. we must consider words beginning wth A,C,H,I,N,S in order
case1- number starting with A
now we will arrange CHINS
in 5! ways =120
case2-NUMBER STARTING WITH C
same we will arrange AHINS
so 5! ways
case3- NUMBER STARTING WITH I
5!
case4- number starting with N
5!
case5 -
NOW NUMBER STARTING WITH S and SACHIN will appear in this list
SACHIN is first word in list of words beginning with S
so 5!+5!+5!+5!+5!+1=601

Find the rank of the word DASMESH

First consider DSMESH, 6!/2! ways=360
DAE (arrange remaining in 4!/2!) = 12
DAH (4!/2!) = 12
DAM (4!/2!) = 12
DASE (3!) = 6
DASH (3!) = 6
DASMEHS
DASMESH
so 360+50=410

Distribute 10 similar apple to A,B,C such that A ,B,C will not get less than 2

A+B+C=10
now our moto in these type of questions to convert it in natural number or whole number form
Like A+B+C=10
we know that min value of A , B and C is 2
so let A=a+2 ((where a > =0)
B=b+2(b > =0)
C=c+2((c > =0))
so a+2+b+2+c+2=10
so a+b+c=4
now our question is to distribule 4 items to 3 persons such that a,b,c could get 0,1,2,,3,4
so number of ways=n+r-1Cr-1
=4+3-1c3-1
6c2

Concept : DISTRIBUTION OF IDENTICAL ITEMS IN DIFFERENT GROUPS

let we have 30 identical mangoes and we want to distribute it to 3 person A,B and C
a) when anyone could get any-number of mangoes ((including zero))

so A+B+C=30
case1- when A=0 , B+C=30
B=0,C=30
B=1,C=29
.
.
.
B=30 ,C=0
so total 31 cases when A=0
case2- Now A=1
so B+C=29 so there will be 30 cases
.
.
same for A=2 ,3 and so on
when A=30 then B+C=0
then B=0 and C=0 so 1 case
so total number of ways=1+2+3.......31
so 31 * 32/2=16 * 31 ways to distribute those mangoes to A,B,C

NOW DIRECT FORMULA
a1+a2+a3.....ar=n
whole number of solution or non integral number of solutions
=n+r-1C(r-1))
as in this case
A+B+C=30
here n=30 ,r=3
so number of ways=n+r-1cr-1
=30+3-1c3-1
=32c2=
32 * 31/2 = 16 * 31

NOW If i put any restriction here like minimum number of mangoes given to A,B,C will be 1
so A+B+C=30
A=1 , B+C=29
so B=1,C=28
B=2,C=27
B=3,C=26
.
.
B=28, C=1
so total =28 cases
,
.
.
now till A=28
B+C=2
so B=1 , C=1
so total ways =1+2+3.....28
so 28 * 29/2 = 14 * 29

NOW DIRECT FORMULA
Natural number of ways to distribute r things to n similar things
n-1Cr-1
here A+B+C=30
so 30-1c3-1 = 29c2 = 29 * 28/2 = 29 * 14

CONCEPT

(1) SUM OF ALL NUMBERS THAT CAN BE FORMED WITH THE DIGITS 2,3,4,5 taken all at a time .. ((means repetition is not allowed))
(2) when repetition allowed

Approach 1)
The total number of numbers formed with the digits 2,3,4,5 taken all at a time = 4!=24
to find the sum of these 24 numbers we will find sum of digits at unit, ten, hundred and thousand places in all these numbers ...
consider the digits in the unit's places in all these numbers
Each of the digit will occur 3! =6 times in the unit place
So total for the digits in unit's place in all the numbers = (2 + 3 + 4 +5 ) * 6
so sum of all numbers = (2 + 3 + 4 + 5) * 3! * ( 10^0 + 10^1 + 10^2 + 10^3))
so ((2+3+4+5) * 3! * (10^4-1)/9
= (2+3+4+5) * 3! * (1111)

DIRECT FORMULA
SUM OF numbers formed by n non zero digits =
((sum of digits)) * (n-1!)) * (11111....n times))
2- total number of numbers in case of repetition=4 * 4 * 4 * 4 = 4^4
SO EACH DIGIT will occur at 4^4/4=4^3 =64 times
so ((sum of digits)) * 64 * (1111)) so formula =sum of digits*n^n * (1111)

CONCEPT

FIND SUM of all numbers greater than 10000 formed by using digits 0,2,4,6,8, no digit repeated in any number

SUM=sum of numbers formed by using digits (0,2,4,6,8)- sum of numbers formed by using digits(2,4,6,8))
((0+2+4+6+8) * (5-1!) * (11111)-(2+4+6+8) * (4-1!) * (1111))

Ordered & Unordered

Find ordered number of positive solutions of abc=30

method-1

abc=540
30=2^2 * 3^3 * 5^1
a^x * b^y * c^z=2^2 * 3^3 * 5^1
a will some number which will contain 2 ,3 and 5
same for b
and same for
and there will be some power of 2 ,3 and 5 in them whatever will be the power 0,1,2 or 3
but 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 1
so x + y + z = 2 (where x,y,z could vary from 0 to2)
so number of ways=2+3-1c3-1=4c2
x+y+z=3
3+3-1c3-1=5c2
x+y+z=1
so 1+3-1c3-1=3c2
so total number of positive solutions will be
=4c2 * 5c2 * 3c2=180
so ordered number of positive solutions will be 180

now if we want total number of integral solutions then we will find negative solutions also
for negative out of a,b,c ,2 numbers will be negative
so choose 2 numbers out of those 3 numbers
so 3c2* positive solutions
so 3c2 * 180
so total ordered solutions =180+3c2 * 180
abc=300
abc=2^2 * 3 * 5^2

so x+y+z=2 so 4c2
x+y+z=1 so 3c2
x+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 * 5^2
not possible
so 6x+y+z=108
so x=108-12/6
so x=16
and y=4
so unordered =x+y=20

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

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

In how many ways can 5^17 be written as a product of three positive numbers
a+b+c = 17
19C2 = 171
when two are same :
2a+c = 17
total 9 solutions
(171-3 * 9)/ 3! + 9 = 33

No need of making cases!
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

|x * y * z| =210.find number of Integral Solutions

210= 2 * 3 * 5 * 7 ..
So positive = 3 * 3 * 3 * 3 = 3^4
Now each of three can be either positive or negative
So 2 * 2 * 2 * 3^4 = 8 * 3^4
Unordered:81 - 3 / 6 +1 = 14

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

In how many ways 1001^2 can be written as product of 3 integers
total ordered and unordered

a * b * c = 7^2 * 11^2 * 13^2
x+y+z=2=4c2=6
x+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 * x
7^2 * x
11^2 * x
13^2 * 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 +16

In how many ways 1000 can b expressed as the product of 3 integers?
ordered and unordered

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

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