Latest Elitmus Aptitude Question SOLUTION: How many 6 digit no can be formed by 0,1,2,3,4,5 so that no is divisible by unit place digit

take 5 at unit place the _ _ _ _ _ 5 thn it is 4*4! and nw take 4 thn it have two cases _ _ _ _ 0 4 and _ _ _ _ 2 4 thn it is 4*3!+3*3! and again for 3 _ _ _ _ _ 3 thn it is 4*4! and nw for 2 it is _ _ _ _ _ 2 so 4*4! and similarly for 1 4*4! so we add oll (4*4!)+(4*3!+3*3!)+(4*4!)+(4*4!)+(4*4!)= 96+(24+18)+96+96+96= 426

fix unit digit :-
for 1 : 96 ways
for 2: 96 ways
for 3: 96 ways
for 4:case 1: when last two digit 04 then 24 ways
case 2: when last two digit 24 then 18 ways
for 5: 96 ways

so all the possible number can formed = 96*4+(24+18) = 426.

total=522
1._ _ _ _ _ 0=5*4*3*2*1=120
2._ _ _ _ _ 1=4*4*3*2*1=96
3._ _ _ _ _ 2=4*4*3*2*1=96
4._ _ _ _ _ 3=4*4*3*2*1=96
5._ _ _ _ _ 4=3*3*2*1=18
6._ _ _ _ _ 5=96
total=522

Ans is 546
0=120
1=4*4!
2=4*4!
3=4*4!
4=4!+3*3!
5=4*4!

408
fix last digit-
'1'---- 96 ways
'2'---- 96 ways
'3'---- 96 ways
'4'---- 24 ways
'5'---- 96 ways

to be divisible by zero the number should end in zero.. for that there are 120 possibilities

Case 1(When last digit is 5):

_ _ _ _ _ 5 and 1st digit should not be 0 so fill 1st place 4type and 2nd place has also 4types and 3rd place as 3types 4thplace has 2types so total will be=4*(4*3*2)=4*4!

Case 2(Last digit is 4):
in this we have two sub case (last two digits should be either 04 or 24)
Sub case 1(last two digits are 04):
_ _ _ _ 04 so 1st place can be filled in 4types,2nd place can be filled in 3types,3rd place can be filled in 2types,4th place can be filled one type so total ways=4!
Sub case 2(last two digits are 24)
_ _ _ _24 so first place can be filled in 3 ways,2nd place can be filled in 3types,3rd place can be filled in 2types,4th place can be filled one type so total ways=3*3!

Case 3(Last digit is 3):
as sum of all digit is 15 so any no. will be divisible by 3
_ _ _ _ _3 and 1st digit should not be 0 so fill 1st place 4type and 2nd place has also 4types and 3rd place as 3types 4thplace has 2types so total will be=4*(4*3*2)=4*4!

Case 4(Last digit is 2):
as all no. end with 2 will be divisible by 2
_ _ _ _ _2 and 1st digit should not be 0 so fill 1st place 4type and 2nd place has also 4types and 3rd place as 3types 4thplace has 2types so total will be=4*(4*3*2)=4*4!

Case 5(Last digit is 1):
as all no. end with 1 will be divisible by 1
_ _ _ _ _1 and 1st digit should not be 0 so fill 1st place 4type and 2nd place has also 4types and 3rd place as 3types 4thplace has 2types so total will be=4*(4*3*2)=4*4!

Repetation alowed or not

(432)
5 >> 96
4 >> 4*3*2*1*2=48
3 >> 96
2 >> 96
1 >> 96
0 >> 0

For 0 at unit place:0 cases
For 1 at unit place:4*4*3*2*1*1 cases=96 cases
For 2 at unit place:4*4*3*2*1*1 cases=96 cases
For 3 at unit place:4*4*3*2*1*1 cases=96 cases
For 4 at unit place:
CASE1:for 04 as last two digits=4*3*2*1*1*1 cases=24 cases
CASE2:for 24 as last two digits=3*3*2*1*1*1 cases=18 cases
total=42 cases
For 5 at unit place:4*4*3*2*1*1 cases=96 cases
Answer=426

Ans: for 0-5!=120
1-No one
2-No one
3-No one(1+2+3+4+5=15)
4-3*(3*3!)=54(-,-,-,-,1/3/5,4)
5-No one
Ans=120+54=174

Thnx for this questions

0,1,2,3,4,5
TOTAL NO 6
SO 6!=720
But in this 720 ways ther exists a numbers starting by 0 012345
012564
like so ways -----
5!=120
all the numbers which started by 0 are'nt 6 digit
so 720-120
-----------------answer---600 6 digit numbers

how 4*3!+3*3! become

(420)
5 >> 96
4 >> 3*3*2*2*1=36
3 >> 96
2 >> 96
1 >> 96
so total numbers are 420

_ _ _ _ _ _ 0 ----> 120 ways
_ _ _ _ _ _ 1 -----> 96 ways
_ _ _ _ _ _ 2 -----> 96 ways
_ _ _ _ _ _ 3 -----> 96 ways
_ _ _ _ _ _ 4 -----> 48 ways (last two digits must be 04 or 24)
_ _ _ _ _ _ 5 -----> 96 ways

total ways = 552 ways