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?

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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 (11 years ago)

Solved anusha Read Solution (13)

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Read Solution (13) :

anusha  (11 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  (11 years ago)

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

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

Richi Sharma  (11 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  (11 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  (11 years ago)

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

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

Ganesh Patil  (11 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  (11 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  (11 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  (11 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  (11 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  (11 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

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sailee  (11 years ago)

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

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DHAKAD NEETENDRA SINGH  (11 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...

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