Latest TCS Aptitude Question SOLUTION: In how many ways you can put 4 letters in 2 post boxes. Any no of letter can be posted in box. a)8 b)4C2 c)24 d) 2P4

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In how many ways you can put 4 letters in 2 post boxes. Any no of letter can be posted in box.
a)8
b)4C2
c)24
d) 2P4

Asked In TCS Nikita (13 years ago)

Solved Vadivelan M Read Solution (10)

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Read Solution (10) :

Vadivelan M  (13 years ago)

2 power 4 is the answer. d)2P4 if P is power.
Reason :
each letter can be posted in any of 2 post boxes. So its 2*2*2*2 = 16

Is this solution Helpfull? Yes (10) | No (4)  

vibhor jain  (13 years ago)

since any number of letters can be posted in box
hence number of permutation of 4 different letters to put in 2 boxes=
n^r
here n=4 and r=2
so 4^2= 4*4= 16.

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

ISITA PRAMANIK  (13 years ago)

8 ways..
1st letter in 2 ways+2nd letter in 2 ways+3rd letter in 2 ways+4th letter in 2 ways=8 ways

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Koushik Paul  (13 years ago)

4C2. Because there are 4 letters and 2 post boxes,so the answer should be 4C2

Is this solution Helpfull? Yes (3) | No (6)  

mohit  (13 years ago)

its definately 8
4 letters can be placed in 2 boxes as 4*3=12 ways.
for selection of box their are 2 ways.
So total ways=12*2=24 ways..

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anil  (13 years ago)

To find the perfect solution follow this link to download the text file

http://uploading.com/files/e68e2fba/tcs2.txt/,tcs2.txt,357 bytes,,Nov 16, 2011,

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

shubham maurya  (12 years ago)

we can put every letter by 4 ways.first letter can be post by 4 ways as well as second letter.
hence total no. of ways=4*4

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sreejesh rajan  (13 years ago)

4c2..... because we want to select 4 letter to the boxes

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anubhav mehra  (13 years ago)

8 ways because 4+4 =8 ways

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hrithik  (12 years ago)

4c2
ways Combinations are dealt with by using factorials...The easiest way to explain is to start small

lets say you have 3 numbers...3, 2, 1. You can only arrange them in groups of 2 (in your question, this would be a 2 number box).

The combinations are as follows

3, 2
3, 1
2, 3
2, 1
1, 3
1, 2

There are duplicates but that was not a criteria, the question was how many ways, 3, 1 is organized in a different way than 1, 3.

Anyway, there are 6 combinations. The best way to explain how the factorial works here is to say you have 3 numbers, but you can only use 2...so you will go back 2 places on the factorial (A factorial is a number multiplied by one integer less, and then again by 1 less, until arriving at 1)

3! = 3 x 2 x 1. But because you can only use 2 numbers, this becomes 3!/1! = (3 x 2 x 1)/1 = 6.

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