i simple dont understand why i get these dbts at this time of yr....

5 ladies and 5 gents are to be arranged in a row..Find no. of ways so that no two ladies are together

Method 1
place 5 gents on 5 seats(leaving one space) with 5! ways and then we have 6! ways to place ladies in vacant seats
ans=6!.5!

Method 2
Total ways to arrange 10 people =10!

Taking all ladies together ,no. of ways=5!.6!

so ans is 10!-(6!.5!)

which method is correct ????????

and ya ...ans is 6!.5!....but atleast tell me my mitskae plzzz

19 Answers

1
madhumitha harishankar ·

yes I agree with u, that is why answer given is right, and method is the first one

24
eureka123 ·

thx everyone [1]

4
UTTARA ·

No Sorry

Ure correct Asish I was getting 8 I mistook 33 2 as 23 3

1
madhumitha harishankar ·

Uttara, can u explain why ashish's answer is wrong?

1
Amber ·

2nd method is wrong .
in 2nd method you have to subtract cases of 2,3 & 4 ladies together.

106
Asish Mahapatra ·

@uttara.. method 1 is definitely correct..

Q2. can u illustrate the mistake in My solution..

I cant seem to find it.. (perhaps due to boards tension..)

4
UTTARA ·

Ya Even I made the same mistake as eureka

Ya u people r correct

So ans is 1st method

4
UTTARA ·

Asish u made a calc mistake

Ans is coming out to be 8

Check ur last step !!

106
Asish Mahapatra ·

For Q1. second method is wrong.

reason is madhu's post given in #12

Q2. 24 = 23.31

Find exponentr of 2 and 3 in 25!
25! = 222.310I (I is integer) = 221.37.2.33*I
= 87*27*2*I

So, ans is 7

1
madhumitha harishankar ·

the 2nd method is wrong because, it makes sure that 5 ladies arent together..but it does not make sure that 2,3,4 ladies arent together

24
eureka123 ·

Q2 whats max power of 24 by which 25! is divisble

1
madhumitha harishankar ·

@ Uttara,
yeah method one also includes case L G G L G L G L G L....but this is not a contradiction, as it still satisfies the condition that no two ladies should be together....and so method 1 is right.

4
UTTARA ·

NO Madhu eureka's Qs says that no 2 ladies r together

So the ladies r to be separated by 1 or 2 or ... gents such that they r never together

1
madhumitha harishankar ·

in Q1, the 2nd method is wrong because it does not account for the cases in which less than 5 but more than 1 ladies are together.

24
eureka123 ·

hmm....still wondering abt Q1 becoz both ways it seems right..

for Q2 i dont have ans...but if u r 100% confident with ur method ,,theni accept ur soln...otherwise we can wait for a someone else to chip in with some for facts

4
UTTARA ·

@eureka :

Ya So I say that ans must have been 2

Wonder y its given the other way [7]

4
UTTARA ·

This is not very logical but

25 ! ---> 2 x 12 ;

3 x 8 ;

4 x 6 ;

24 ;

3 x 4 x 2 (from 9 14 16) ;

3 x 4 x 2 (from 9 16 18 ) ;

3 x 4 x 2 (from 15 16 22) ;

SO 7 ??

24
eureka123 ·

so whats the problem here ??
L G G L G L G L G L

no 2 ladies together condition is staisfied...

4
UTTARA ·

Method wise 2 seems correct

Becz In Method 1 Apart from those alternating cases u may have one case this way too

L G G L G L G L G L

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