Latest Infosys Aptitude Question SOLUTION: Q. {1,2,3,4,5,5,6,6}how many four digited numbers can be formed without repiting the numbers?

• Placement Papers
• /
• Infosys
• /
• Numerical Ability
• /
• Permutation and Combination

Infosys Question Papers: Infosys Interview Experience(10) INFOSYS Company Profile

(#M40031979) INFOSYS Numerical Ability question logical reasoning Keep an EYE

Q. {1,2,3,4,5,5,6,6}how many four digited numbers can be formed without repiting the numbers?

Asked In Infosys sathvika (10 years ago)

Solved anusha Read Solution (13)

Is this Puzzle helpful?   (5)   (0) Submit Your Solution Permutation and Combination

Read Solution (13) :

anusha  (10 years ago)

n*(n-1)*(n-2)...
hence four digits with out repetition total 6 digits=n
6*5*4*3=360

Is this solution Helpfull? Yes (21) | No (5)  

bakshu  (10 years ago)

8*7*6*5/2*2=420

Is this solution Helpfull? Yes (11) | No (2)  

Richi Sharma  (10 years ago)

we have to form 4 digit number
_ _ _ _
8*7*6*5
as first place can hold any of the digits 1,2,3,4,5,5,6,6
second place can hold 7 digits
third place can hold 6 digits
and fourth place can hold 5 digits
since digit 5 is repeated 2 times and digit 6 is repeated 2 times

required number of 4 digit number = (8*7*6*5)/2!*2! = 420

Is this solution Helpfull? Yes (10) | No   

Satyaranjan  (10 years ago)

Updated:

1.nos containing only 1,2,3,4,5 and 6:6C4x4!=360
2.nos containing two 5's(Ex:5512):(5C2x4!)/2!=120
3.nos containing two 6's(Ex:6612):(5C2x4!)/2!=120
4.nos containing two 5's and 6's(Ex:5566):(4!)/2!x2!=6
Total nos:360+240+6=606

Is this solution Helpfull? Yes (8) | No (1)  

T G  (10 years ago)

8P4/2!*2! AS THERE ARE 2 5S AND 2 6S

Is this solution Helpfull? Yes (7) | No (3)  

Ganesh Patil  (10 years ago)

ans=420
arrangement of "n" object at "r" place where "r1" and "r2" are of common type
=[n!/(n-r)!]*(1/r1!*r2!.....rn!)

here
n=8;
r=4;
r1=2;
r2=2;
ans=(8!/4!)(1/2!*2!);
=(8*7*6*5)(1/4)
=2*7*6*5
=420

Is this solution Helpfull? Yes (4) | No   

rohit  (10 years ago)

guys just for a second imagine that there are only 1 5 and 1 6 so if we have 1,2,3,4,5,6 then for 4 digit no we can make following adjustment without any repetition
for 1 st position we have 6 choice
for 2 nd position we have 5 choice
for 3 rd position we have 4 and for 4th we have 3....so no of possibility are 6*5*4*3...and now come back to asked question in which we have 2 5s and 2 6s...
so final answer is 6*5*4*3/2!*2!

Is this solution Helpfull? Yes (2) | No   

sdfgvdhgf  (10 years ago)

nos containing only 1,2,3,4,5 and 6:6C4x4!=360

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

DHAKAD NEETENDRA SINGH  (10 years ago)

total four digit no's are
according to fundamental principle of counting
four digit no= (8*7*6*5)/2!2!
= 720 ans...

Is this solution Helpfull? Yes | No (6)  

Satyaranjan  (10 years ago)

1.nos containing only 1,2,3,4,5 and 6:6C4x4!=360
2.nos containing two 5's(Ex:5512):(5C2x4!)/2!=24
3.nos containing two 6's(Ex:6612):(5C2x4!)/2!=24
4.nos containing two 5's and 6's(Ex:5566):(4!)/2!x2!=6
Total nos:360+48+6=414
@sathvika:Is it right?

Is this solution Helpfull? Yes | No (4)  

Abhishek Bhardwaj  (10 years ago)

8c4/2!*2!
Here for choosing 4 we have 8c4
But for repetion divide with 2! for 5 and 6 value are repited

Is this solution Helpfull? Yes | No (4)  

sailee  (10 years ago)

are these Qs sufficent to crack the aptitude test of infosys?

Is this solution Helpfull? Yes | No   

DHAKAD NEETENDRA SINGH  (10 years ago)

right answer for this question
1,2,3,4,5,5,6,6
4 different(1,2,3,4) 2 pair(5,5 6,6)
I. both pair = 4!/2!2! = 6
II. 1 pair and 2 different = 2c1*4c2*4!/2! = 288
III. all different = 4! = 24
total ways-
= 24+288+6
= 318 ans...

Is this solution Helpfull? Yes | No   

Please Login to Read Solution.