MAths DOubt

[The Villain]

 Q1 ..In how many ways can one or more of 5 letters be posted in 4 mail boxes if any letter can be posted in any of the boxes?
a 5^4
b 4^5
c 5^5-1
d 4^5-1

Q2..In how many ways can 5 prizes be given to 3 boys when each boyvis elgible for 1 or more prizes?
a 5^3
b 3^5
c 5P3
d 5C3

[Homer Simpson]

1(b), 2(b) - wrote a by mistake!

[The Villain]

The ans are 1 c and 2 b.. How??

[SINGHAM]

Q1 ..In how many ways can one or more of 5 letters be posted in 4 mail boxes if any letter can be posted in any of the boxes?

ANS: "one or more" is critical here.

Suppose he posts only 1 letter. So, which one is that letter? There are 5C1 ways to choose that letter. Now let we post it. That letter can be posted in 4 different letterboxes.

Suppose he posts only 2 letter. So, which one is that letter? There are 5C2 ways to choose that letter. Now let we post it. That letter can be posted in 4^2 different ways.  How did we get 4^2 ? We can post first letter in 4 ways, and without regard where it was posted, we can post second letter in any one of the 4 boxes.

The process goes on until,

Suppose he posts 5 letter. So, which one is that letter? There are 5C5 ways to choose that letter. Now let we post it. That letter can be posted in 4^5 different ways.  How did we get 4^5 ? We can post first letter in 4 ways, and without regard where it was posted, we can post second letter in any one of the 4 boxes, and without regard where first was posted, we can post the third in 4 ways, and so on, upto 5 letters.

So, the answer is 5C1*4+5C2*4^2+...+5C5*4^5=(1+4)^5-1


@Ron, what is  meant by one or more prizes in Q2? Note that 3^5 can't be the answer, as it also allows that all prizes given to boy 1, etc, which is ruled out in question itself!

Also, where did you get these question?

[The Villain]

Singham...thanxx man!!
I have a book...got it fr there...

[SINGHAM]

Ok. Will you please let me know that book, please? I will be obliged.

[The Villain]

Mba material of Time institute...!!!I guess permutations is permutations...do it fr anywhere ryt

[Shefali]

@singham...hw did u simplify it to get the ans

[SINGHAM]

O ya! That's true!

[SINGHAM]

Pardon, what simplification are you talking about?

[Shefali]

5C1*4+5C2*4^2+...+5C5*4^5
hw did u simplify this??

[Shefali]

Singham pls explain

[SINGHAM]

O ya! It is Binomial expansion, u know!

Note that 5^5=(1+4)^5=5C0+5C1*4+5C2*4^2+5C3*4^3+5C4*4^4+5C5*4^5
Now compare with the expression 5C1*4+5C2*4^2+...+5C5*4^5 .
So, 5C1*4+5C2*4^2+...+5C5*4^5 = (1+4)^5-5C0= (1+4)^5-1