Latest Elitmus Aptitude Question SOLUTION: The no of digits greater than 6 * 10^6 that can be formed by 0,5,5,6,6,9,9 is: OPTIONS a)320 b)540 c)720 d)dont remember This question was asked on 22nd feb ph test

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(#M40119488) ELITMUS Numerical Ability question Permutation and Combination Keep an EYE

The no of digits greater than 6 * 10^6 that can be formed by 0,5,5,6,6,9,9 is:
OPTIONS
a)320
b)540
c)720
d)dont remember

This question was asked on 22nd feb ph test

Asked In Elitmus sheikh shahrukh (9 years ago)

Solved ASHISH MUKHERJEE Read Solution (8)

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Read Solution (8) :

kartheek  (9 years ago)

I think the question is how many 7 digit nos can be formed using digits 0,5,5,6,6,9,9 which are greater than 6(10^6) i.e., 6000000.
Let the 7-digit nos be _ _ _ _ _ _ _
To get a number greater than 6(10^6) , we can fill the first blank with either 6 or 9.
Case 1:
The first digit is filled with 6 and the remaining digits can be filled in 6!/(2! *2! ) ways {because it is a permutation of 6 objects out of which 2 objects are similar (5,5) ,another 2 objects are similar(9,9) and the remaining are different (0,6) }
Therefore no of such 7 digit nos which start with 6 are 720/4 =180

case 2:
similarly we can have the first digit as 9 and remaining 6 digits are permuted.
In that case no of such nos is also 6!/(2! 2!) = 180

So total nos that can be formed which are greater than 6(10^6) = 180+180 = 360
Answer is 360.

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lokesh  (9 years ago)

6_ _ _ _ _ _ here empty can be use 055699 that can be arranged 6!/2! 2! =180
9_ _ _ _ _ _ can also used 055669 that can be arranged 6!/2! 2!=180
so total number are 180+180=360

Is this solution Helpfull? Yes (21) | No   

ASHISH MUKHERJEE  (9 years ago) Best Solution

case 1: when left most side will fill with 6 then
2*6*5*4*3*2*1/2!*2!*2! = 180
case 2: when left most side will fill with 9 then
2*6*5*4*3*2*1/2!*2!*2! = 180
so, total no formed = 180+180 = 360

Is this solution Helpfull? Yes (10) | No (1)  

Ramesh Layek  (9 years ago)

(7! / 2!*2!*2!) - (6! / 2!*2!) - (6! / 2!*2!*2!)
=630-180-90
=360

Is this solution Helpfull? Yes (2) | No (1)  

Sumaiya  (9 years ago)

for the number to be greater than 6*10^6 i.e 6000000, the left most number should be either 6 or 9.
like for example: 6055699 or 9055669
consider the first set of number which begins with 6. In that we have the left most number fixed and the rest 6 digits that can be varied.
so its 6*5*4*3*2*1 = 720.
but in that numbers we can see that 5 is repeated twice and 9 is repeated twice, hence we need to divide by 2! twice: the actual solution will be (6*5*4*3*2*1)/(2! *2!) which is equal to 180.
this 180 is the possibilities with starting digit as 6.

the same has to be done for 9 as well..
even in that numbers the starting number is fixed and the rest 6 digits are variable.
again here 5 is repeated twice , and 6 is repeated twice. hence the solution will be
(6*5*4*3*2*1)/ (2! * 2!) = 180.

total we have 180+180=360
hence the answer is 360

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mukesh pathak  (9 years ago)

4*6!/2!*2!*2!

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kumari anjana  (8 years ago)

(4*6*5*4*3*2*1)/(2*2*2)=360
ans (d)

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DOLLY  (4 years ago)

total available digits = 7 (0,5,5,6,6,9,9)
__ __ __ __ __ __ __
places : 1st 2nd 3rd 4th 5th 6th 7th

for 1st position possible numbers are (6,6,9,9) so 4
because of repetition 4/(2! * 2!)

for places 2nd to 7th
6*5*4*2*1
because of repetition remaining number i.e. (5,5) = (6*5*4*2*1)/2!

altogether 4* (6*5*4*2*1) / (2! * 2! * 2!) = 360

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