Combinatorics

Question

How many 4 digit numbers divisible by 5 can be formed with the digits 0,1,2,3,4,5,6 and 6?

a) 220

b) 249

c) 432

d) 288

Comments

Two cases I have assumed :
I) Unit digit 0 : 6*5*4 = 120
    Unit digit 5   5*5*4=  100
Adding both of these =220
II) In the above case i didn`t care about two 6`s.
Lets do that 
Again 
    Unit digit 0 : 5+5+5=15
    Unit digit 5   5+5+4=14 In this case we can`t have 0 in MSD
Adding both of these =29
Complete=249. Though i think i covered all cases but lets see what is the answer.!

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yes, it will be 249

i was doing some silly mistake..

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Yes, 249 is the correct answer.

Answer 1

Number will be divisible by 5 , when last digit is eight 0 or 5.

CASE 1)  last digit is 0

Now as there are two 6's , so we again have to make 3 cases under case 1

case a ) number doesn't contain any 6's

case b ) number contains only one 6

case c ) number contains both 6's

CASE 2) Last digit is 5

again there also three cases just like case 1.

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Now logic part is done , only calculation remains.

CASE 1  :

part a) 5*4*3=60 ways

part b) as this time we have one 6 , that can be at 1's place or 2nd or 3rd => 3 ways to place 6 and other digits 5*4 total = 20*3=60

part c) this time both 6 present , first place those two 6's to those available three places in 3C2 ways

total ways =3*5=15

case 1 total = 135

CASE 2 :

part a)  4*4*3=48 ways

part b ) 5*4+4*4+4*4=52 ways

part c) 5+5+4=14 ways

total of CASE 2 =114

total = case 1 + case 2 = 249 ways

Comments

Thanks, Rupendra, for b) & c) in CASE 2 a one-line description would be helpful like you did in CASE 1.

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CASE 2 - part b)

last position is fix with 5 , so to fill last digit , one way

scene 1) when 6 is at first position , now for digit two there are 5 possibilities either 1 comes or 2 or 3 or 4 or 0 now after placing one of those five at second place , at third place there are 4 possibilities as repetition is not allowed. so 5*4 choices,

scene 2) 6 is at second place. and 5 at last. at first place 0 can't come(we have to make four digit number) so or first place 4 choices and third too 4 choices(as 0 can be place here).so 4*4 choices

scene 3) 6 is at third place and 5 at last. to fill first place 4 choices and same for second place so 4*4 choices.

Part c)

scene 1) 6 is at first and second place , 5 is at last. now to fill third place 5 choices(0,1,2,3,4)

scene 2) 6 is at first and third place , again 5 choices

scene 3) 6 is at second and third place. to fill first place 4 choices(as 0 can't be place at first place)

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Thanks a lot, Rupendra!