Latest TCS Aptitude Question SOLUTION: How many words can be formed by using 4 letters at a time of word "SURPRISE" m4maths

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How many words can be formed by using 4 letters at a time of word "SURPRISE" m4maths

Asked In TCS palane (10 years ago)

Unsolved Read Solution (11)

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ABHAY KUMAR  (10 years ago)

the answer must be 420. it should be calculated like this:-
8p4/2!2!=420

Is this solution Helpfull? Yes (18) | No (7)  

Jagadeesh Rampati  (10 years ago)

given word SURPRISE
first make repeated letters a set RR,SS two sets remaining are U,I,P,E
now
case 1:
4 letter word with all(two sets) repeated letters= 2c2* 4!/2!2!=6
case 2:
4 letter word with one set repeated letters and remaining=2c1*5c2*4!/2!2!=120
now with another set=120
total in this case=240
case 3:
4 letter word with no repetition=6c4*4!=360
total no of words=6+240+360=606

Is this solution Helpfull? Yes (17) | No (11)  

karthigabalamurugan  (10 years ago)

OMG!!!! different solutions....wich one is correct..admin pls help us...

Is this solution Helpfull? Yes (6) | No (1)  

lokesh  (10 years ago)

8!/2!*2!=40320/4=10080

Is this solution Helpfull? Yes (4) | No (9)  

Tripti Singh  (10 years ago)

8*7*6*5=1680

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

Mitali  (10 years ago)

Break surprise in to 3 groups --> (UPIE) , (RR) ,(SS) .
We can take out four letters in any of the following ways :
1) take all four form first group and arrange them -> 4! ways OR
2)take 3 from the first group and one from either second or third group (4C3 *2) and arrange them in 4! ways --> 4C3*2*4!=192 ways OR
3) take two from the first group and (4C2) , an R from the second group and an S from the last group (1) and arrange them in 4! ways--> 4C2*1*4! =144 ways OR
4) take two from the first group and (4C2) and either take two R's from the second group or two S's from the last group (2) and arrange them in 4!/2! ways (as two letters will be same this time) --> 4C2*2*4!/2! =144 ways OR
5) Take 1 from the first group (4) and 3 from the second and third groups (2 -Two R from first and one S from third or Two S form the third and one R from the first) and arrange them in 4!/ 2! ways --> 4C1*2*4!/2! =96 ways. OR
6) none from the first group and all four from second and third groups and arrange them in 4!/(2!* 2!) = 6 ways ,
Total number of ways = 24+192+144+144+96+6= 606

Is this solution Helpfull? Yes (2) | No   

palane  (10 years ago)

sorry lokesh , ans is 606 but i dont know how, solution is not clear were is found. please help me with clear solution

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

nagashree  (10 years ago)

four words from 8 words can be formed from 8c4(is formation not arraangement so it cant be permutation)
s and r are twice so 2!*2!
the ans will be 8c4/2!*2!

Is this solution Helpfull? Yes (1) | No (3)  

Rudra  (10 years ago)

plzzzz tell what is the exact ans is???

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

NareshKumar  (10 years ago)

the letters can be spilt into SS,RR,UPIE
arrange non repeating ie UPIE=4!=24
select from UPIE+SS 4 ie =6P4/2!=180
select from UPIE+RR 4 ie =6P4/2!=180
now select SS|RR ie 4!/2!2!=6
total=390

Is this solution Helpfull? Yes | No (2)  

Nitin  (10 years ago)

8C4*[4!/(2!*2!)]=420

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