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(#M40028000) TCS question tcs Keep an EYE

"LEADING" arrange it in such a way that at least two vowels always together...??

Asked In TCS rakesh (11 years ago)

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Prakash kumar  (11 years ago)

first for (AEI)LDNG 5!*3!=720
SECONDLY FOR (AE),(EI),(AI) + REMAINING LETTER 3(6!*2!)= 4320
TOTAL 5040 ARRANGEMENTS

Is this solution Helpfull? Yes (32) | No (9)  

chaitanya  (11 years ago)

The word ‘LEADING’ has 7 different letters.

When the vowles EAI are always together, they can be supposed to form one letter.

Then, we have to arrange the letters LNDG ( EAI ).

Now, 5 (4 + 1 = 5 ) letters can be arranged in 5 = 20 ways.

The vowles ( EAI ) can be arranged among themselves in 3 = 6 ways.

Required number of ways = ( 120 × 6 ) = 720.

Is this solution Helpfull? Yes (9) | No (7)  

Rachit  (11 years ago)

1. when all three are togethr
5! * 3! = 120*6
2. when any 2 are together from E , A , I
6! * 2! * 3C2(choosing any 2 from given 3)= 1440*3
so final = (1440*3)+(720).

Is this solution Helpfull? Yes (8) | No (2)  

Saurav Mandholiwal  (11 years ago)

There are 2 possiblities i.e.
1) Either all the three vowels will come together
2) Or 2 vowels will comes together

for 1st condition LDNG "EAI" = LDNG with "EAI" can arrange in !5 ways and then "EAI" can arrange in !3 ways therefore !5 * !3 = 120*6=720

Now same for the second 1 LDNGI "EA" can arrange in !6 possible ways but "EA" only can arrange in !2 ways and therefore !6*!2= 720* 2= 1440

Now the total no. of ways of arranging these 2 = 1440+720= "2160"

Is this solution Helpfull? Yes (7) | No (9)  

Guru Pranav  (11 years ago)

THERE WILL BE TWO CASES:-
CASE 1:- 2 VOWELS WILL BE TOGETHER 2*6!=1440
CASE 2:- 3 VOWELS WILL BE TOGETHER 5!*3!=720

1440+720= 2160 ANS.

Is this solution Helpfull? Yes (5) | No (2)  

pritish beura  (11 years ago)

there are 7 ltrs...out of which we have to keep together A E I....we can keep it in 3! ways...now we can take aei & it's other format as one ltr so we have now 5 ltr....we can arrange it with 5! ways..
so the total ways= 5!*4*
=720 (ANS)

Is this solution Helpfull? Yes (1) | No (5)  

Rajeev kumar  (11 years ago)

Total no. of words using L,D,N,G and A,E,I will be 7! i.e.,5040.
5040 contains three kind of words.
1st = Three vowel together(like LAEIDNG,AEILDNG, etc)=720
2nd = Two vowel together (like LAEDNGI,LAIDNGE, etc)=2880
3rd = No vowel together (like LADENIG, ALEDNIG, etc)=1440

Now,no. of ways when no vowels are together can be calculated from below
*L*D*N*G* i.e.,5P3*4!=1440
Hence,no. of ways to have atleast two vowels together=5040-1440=3600
Required answer should be 3600

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

Payal  (11 years ago)

for (AEI)LDNG it is 5!=120
for (AEI)= 3
120*3=360

360 is the answer

Is this solution Helpfull? Yes | No (6)  

Arnab Saha  (11 years ago)

L(EA)DING=6 letters which acn b arranged in 6! ways..
which is eqaul to 720 ways..

Is this solution Helpfull? Yes | No (3)  

PRAVEEN K H  (11 years ago)

This is 100% Correct Solution

1) First take (AEI)+LDNG => 3! * 5 ! = 720
2) (AE)+ILDNG => 2! * 6! = 1400 ( but this includes the possibility of AEI also )
So 1440-720=720
3) Similarly for EA and IE in case 2 we get 720 and 720..

So the final answer will be 720+720+720=2160

Is this solution Helpfull? Yes | No (1)  

priyanka bhowmick  (11 years ago)

6!*3C2=720*3=2160

Is this solution Helpfull? Yes | No (1)  

Pallabi Basu  (10 years ago)

EAI L D N G
3C2*2*6!=4320

Is this solution Helpfull? Yes | No (1)  

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