Ace Test Series: Combinatory - Permutations And Combinations

Question

Comments

natural position means like 1 should not be at first positions and 3 should not be at 3rd..like that??

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Ya like that!!

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@vijay could u plz elaborate this answer..??

Answer 1

Permutations possible, 

when 1 is at 1st position P(1) = 4!. ( no condition on other 4 integers)

when 3 is at 3rd position P(3) = 4!.

when 5 is at 5st position P(5) = 4!.

when 1 and 3 are in 1st and 3rd position respt. P(1,3) = 3!.
(no condition on other 3 integers( 2, 4, 5))

when 1 and 5 are at 1st and 5th position respt. P(1,5) = 3!.

when 3 and 5 are at 3rd and 5th position respt. P(3,5) = 3!

when 1,3, and 5 are at 1st, 3rd and 5th position respt. P(1,2, 3) = 2!.

So, Total possible permutations when at least one among 1, 3, and 5 is at its correct natural position, = 

  P(1 || 3 || 5) = 3*4! - 3*3! + 1*2! = 56.

Hence, required ans = 120 - 56 = 64. (Ans) 

Comments

But here some cases repeating

{when 1 and 5 are at 1st and 5th position respt}$\cap${ when 1,3, and 5 are at 1st, 3rd and 5th position respt.  }. =when 1,3, and 5 are at 1st, 3rd and 5th position respt.

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@Srestha, I used this concept.

I think, your case should also be there in that permutations.

Answer 2

1 , 2 , 3 , 4 , 5

This 5 number can purmuted in 5! ways=120 ways

Now, 1 cannot be placed in 1st, 3 cannot be placed in 3rd, 5 cannot be placed in 5 th.

Now, think all are in their position.

So, remaining 2 element can be placed as 2! ways.=2...................i

Now, think only 1 and 3 in their position.

Other elements can place in 3! ways.......................ii

But we have to remove te case i from case ii

So, here 3!-2! ways.

Now, similar case for 3,5 in their position and 1,5 in their position

So, $3\times (3!-2!)$ways..............iii

 

At last we have to think only 1 in it's position.

so, here we have to remove case i and case ii as duplicate case

Say, for 1 is in place

From case ii we are getting similar case when (1,3) is in position, and when (1,5) is in position and also case i , when all 1,3,5 all are in position.
$4!-(2\times (3!-2!)-2!)-2!$

Similar case if only 3 is in position and only 5 in position

i.e.$(4!-(2\times (3!-2!)-2!)-2!)\times3.$............iv

 

So, remaining case are 120-(2!+$(3!-2!)\times3$+$(4!-(2\times (3!-2!)-2!)-2!)\times3$.)=120-2-12-18=88

Answer 3

5! (total possible)- (3C1){when one odd number is in correct position}*4!(permuting rest of 4 numbers) + (3C2)*3! {we over subtracted 1 and 3 and 5 so adding them back } - (3C3)*2! {when all three are in correct position you have 2 ways to arrange 2 and 4} = 120-72+18-2=64

Comments

what is (3C2)*3!  for?when we have considered the cases when two numbers are at their corect positions?

dun you think we need to add 2*(2!) for the case when all are at correct position because we have already considered these cases 3 times when anyone was at its correct position and since you are subtracting by 2! for the same.

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when i considered 1 to be placed in the correct position , then there may be 3 also which comes in the correct position so there are 3 pairs(1,3) (1,5) (3,5) so these must be added now i subtracted when 1 and 3 comes in their natural positions so we subtracted them (1,3,5 must be added ) so that is the last term