IIT JEE , AIEEE Forum - Mathematics , Algebra

shitij_iitD

*Find the numbers of 6 digits natural nos,where each digits appears at least twice.

* How many numbers from1to1000 are not divisible by 2,3,5?

aditya_arora04

Hi,

Ans. 1 :

Look, for this let us calculate the number of ways in which a 6 digit number can be made without repeating any digit. Then we will subtract it from the total number of ways.

Total number of ways = (10)⁵ x (9) [ In first digit only 1-9 can appear ]

Ways when no digit is repeated = 9 x 9 x 8 x 7 x 6 x 5

Subtract and you get the answer.

ramyadiamond

The no.s that are simulatenously divisible by 2,3,5 are the no.s which are the multiples of 30.

So, the total no. of terms in this sequence,

30,60,................990 which contains no.s that are divisible by all three no.s be n.

990=30+(n-1)30

n=33

So, 33 terms are divisible by 2,3,5 hence

no.s not divisible by them are 1000-33=967

ankur.kkhurana

i don't agree with both of u .HE SAID WHERE EACH DIGIT REPEAT TWICE AT least .So we hav to make cases.
so ans is
case 1
> 2 + 2 + 2
so
9x8x7x 6!/(2! x 2! x 2!)
case 2
> 2 + 4
9x8x6!/(2! x 4 !)

case three
3+3
9x8 x6!/( 3!x 3! )

case 4
>6+0

9
add abv and get answer

nick

well according to me the method is

case1
2+2+2
i.e. the no. can be of form
aabbcc
but we have to see that 0 can be included therefore no. of ways arof selecting the no. are
9*9*8 and these can be arranged in 6C3 ways
hence total are 9*9*8*20=12960

case2
3+3
aaabbb
similarly
9*9*6C2=1215

case3
4+2
aaaabb
no. of ways of selecting the no. is 9*9
but no. of ways of arranging them works out to be 12(i may be wrong here )hence total ways are 972

finally case4
6+0
i.e 9

there total is 15156

plss crt me if i am wrong

ankur.kkhurana

i hav given same answer. if u can see abv

nick

no offence dude,
i did not work out the ans. from ur method

and is our answer right???/

Rishi_08

Here 4 ur second Qs:

Let A being the property of being divisible by2, B the property of being divisible by 3,C the property of being divisible by 5.

Then,

AB=being divisible by 6

BC=being divisible by10

CA=being divisible by 15

ABC=being divisible by 30

Then we have to find n(A'B'C')=n-n(A)-n(B)-n(C)+n(AB)+n(BC)+n(CA)-n(ABC)

We are given that n=1000

So n(A)=[1000/2]=500

n(B)=[1000/3]=333

n(C)=[1000/5]=200

n(AB)=[1000/6]=166

similarly

n(BC)=66

n(CA)=100

n(ABC)=33

Therefore n(A'B'C')=1000-500-333-200+166+100+66-33=266.

Hope u find it useful.

smit.chandra

ans to Qno1 is (9*10*10*10*10*10)-(9*9*8*7*6*5)!!!!!100% sure!

magiclko

_ _ _ _ _ _

a b c d e f

if the digits has to be repeated atleast twice in a six-digit number, then following cases are possible

case 1: 3 digits are choosen and each is repeated twice

then for position a, we have 9 digits to choose from (0 nt included)

and then this digit is at one more place , thrfore total no of ways of choosing that digit and then placing it = 9C1 X 5C1 (5C1 cuz, its already at a position, nd out of rest 5 positions, one has to be choosen)

similarly, for the other digits total no of ways of choosing it and arranging it = 9C1 X 4C2 (9 cuz, 0 can also be thr )

and for the last digit, total no of ways = 8C1 X 2C2

thrfore total no of ways, in which the number can be formed using 3 digits, each repeated twice 9C1 X 5C1 X 9C1 X 4C2 X 8C1 X 2C2

case 2: 2 digits are choosen each repeated thrice

choosing the digit which has to be at "a" = 9C1

and then arranging it two tyms more = 5C2

then again choosing the other digit and arranging it in rest 3places = 9C1 X 3C3

thrfore total no of ways = 9C1 X 5C2 X 9C1 X 3C3

case 3: 2 digits are choosen, one is repeated four tyms, and one 2 tyms

if the digit at "a" is repeated

four tyms, then no of ways of forming the number = 9C1 X 5C3 X 9C1 X 2C2

two tym, then no of ways of forming the number = 9C1 X 5C1 X9C1 X 4C4

case 4: all digits are repeated,

in that case we have to choose just one digit..thrfore total no of

ways of forming such number = 9C1

thfore total no of ways = {9C1 X 5C1 X 9C1 X 4C2 X 8C1 X 2C2} + {9C1 X 5C2 X 9C1 X 3C3} + {9C1 X 5C3 X 9C1 X 2C2} + {9C1 X 5C1 X9C1 X 4C4 } + 9C1

Arjun

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