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Numerical Ability
Permutation and Combination
the letters in the word ROADIE are permuted in all possiible ways and arranged in alphabetical order. find thw word in 44 th rank
Read Solution (Total 9)
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- arrange in alphabetic order
ADEIOR
now
AD - - - - = 4! = 24
AE D _ _ _ = 3! = 6
AE I _ _ _ = 3! = 6
AE o _ _ _ = 3! = 6
till now we got 42 words (24+6+6+6)
now 43 word will be
AERDIO
so 44 will be
AERDOI
- 10 years agoHelpfull: Yes(29) No(11)
- A[DEIOR] =5!=120 NOT POSSIBLE
AD[EIOR] = 4! = 24
AE[DIOR] = 4! = 24
AED[1OR] = 3! = 6+24=30
AEI[DOR] =3! = 6+30 =36
AEO[DIR] = 3! = 36+6 =42
NOW 43RD LETTER WOULD BE AERDIO
AND HENCE 44TH IS AERDOI
- 10 years agoHelpfull: Yes(12) No(0)
- given is ROADIE,need to arrange in this order ADEIOR
so,
to get the 1st letter A (present in 3rd position) =3! (now A_ _ _ _ _)
2nd D(present in 3rd position after arranging A)=3! (AD_ _ _ _)
3rd E(in 4th position after arranging AD)=4! (ADE_ _ _)
4th I( in 3rd position)=3! (ADEI _ _)
5th O(in 2nd position)=2! (ADEIO _)
6th R(in 1st position)=1! (ADEIOR)
so finally 3!+3!+4!+3!+2!+1=44 - 10 years agoHelpfull: Yes(4) No(6)
- @adity bhowmik plz read the question carefully,u misunderstood it.
- 10 years agoHelpfull: Yes(4) No(0)
- AEORID-42TH
AERDIO-43TH
AERIDO-44TH - 10 years agoHelpfull: Yes(1) No(0)
- Solve by checking options....
Suppose we have AERDOI in option
A D R D O I
5! 4! 3! 2! 1! 0!
taking first alphabet A check how many alphabets are smaller than A. Here no alphabet is smaller than A so write zero(0) below A
A D R D O I
5! 4! 3! 2! 1! 0!
0
//ly check the other alphabets
A D R D O I
5! 4! 3! 2! 1! 0!
0 1 3 0 1 0
Ans is: 0*5! + 1*4! + 3*3! + 0*2! + 1*1! + 0*0! = 44
so 44th word is AERDOI - 9 years agoHelpfull: Yes(1) No(2)
AEROID=48th rank
AERODI=47th ..
AERIOD=46..
AERIDO=45..
AERDOI=44 th rank- 10 years agoHelpfull: Yes(0) No(2)
- i cant understand what is meant by rank?can u plz explain it
- 10 years agoHelpfull: Yes(0) No(0)
- clearly explain it
- 9 years agoHelpfull: Yes(0) No(0)
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