Latest Infosys Aptitude Question SOLUTION: if a die has 1 6 and 3 4 and 2 5 opposite each other how many such dies can be made

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if a die has 1 6 and 3 4 and 2 5 opposite each other how many such dies can be made

Asked In Infosys Priyanka Uttam Wankhede (8 years ago)

Unsolved Read Solution (12)

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Read Solution (12) :

saroj  (8 years ago)

ans 12
if one face have 1,3,2, then the crossponding face will be occupied by respective pair i.e (6,4,5).
so there r 3 place and we have 3 place soit can be arranged in 3! way i.e 6 way and every pair will be suffled in 2 way so ans will be6*2=12.

Is this solution Helpfull? Yes (23) | No (2)  

chitra  (8 years ago)

Ans::24 ways

if the dice shows 6 on top then 1 will be at the bottom.

so the remaining 3,4 and 2,5 can be plotted in 4 ways so four different dies possible

if 1 is placed on the top then again 4 ways


4*2+4*2+4*2=24 ways

Is this solution Helpfull? Yes (4) | No (6)  

Dhvani Agola  (8 years ago)

12
there are 3 pairs : (1,6),(3,4),(2,5)
and each pair has 2 sides
so, 3*2*1*2 = 12

Is this solution Helpfull? Yes (4) | No (4)  

Jatin  (8 years ago)

2!+2!+2!=6
all can be arranged in 3! ways
hence 36

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

meher prasad J  (7 years ago)

3!*2*2*2=48

Is this solution Helpfull? Yes (2) | No   

sapna sikarwar  (7 years ago)

whats the correct ans

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shenba  (7 years ago)

permutation and combination

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NISHANT BHARTI  (7 years ago)

Ans: 8
Taking 1 at one of its faces, the other possibilities for other numbers 2,3,4,5 is 4.
Now interchanging 1 and 6 (2 ways), we will have 4 more possibilities.

Hence, total number if ways: 4*2=8

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

Chandana  (5 years ago)

Let's take a blank die which has faces a,b,c,d,e,f

(a,b),(c,d),(e,f) are opposite pair of sides

(1,6),(3,4),(2,5) are 3 possible opposite side sets required

These three can be arranged in above blank cube in 3*2 = 6 ways

Each of these pairs can be arranged in 2 ways (1,6) or (6,1).

Therefore Total number of dice possible = 6*2 = 12

Is this solution Helpfull? Yes (1) | No   

asha ramoliya  (7 years ago)

if take 1 and 6 as fixed then total 8 posiblities of (3,4) and (2,5)
then alter 1 and 6 then again 8 posoblities
same for (3,4) and (2,5)
total 48 posiblities but here some 8 posoblities is repeated so answer is:40

Is this solution Helpfull? Yes | No (1)  

lokesh dwivedi  (7 years ago)

Let us assume their r 2 configurations at a vertice
A. and. A
B. C. C. B


If we interchange A and B in the second configuration then it becomes the first configuration

This means two possible basic arrangements of the the 3 sets at a vertice

Now, 1_6 ,3_4, 5_2. Can interchange.


This gives total ways as 2*2*2*2

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Ru  (6 years ago)

8 ways = 2*2*2
1 6 1 6 1 6 1 6
3 4 3 4 4 3 4 3
2 5 5 2 2 5 5 2

6 1 6 1 6 1 6 1
3 4 4 3 4 3 3 4
2 5 2 5 5 2 5 2

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