Hexaware
Company
Numerical Ability
Permutation and Combination
Using all the letters of the word 'NOKIA' how many words can be formed, which begin with N and end with A?
A)6 B)18 C)12 D)24
Read Solution (Total 7)
-
- There are five letters in the given word.
consider 5 blanks
The first blank and last blank must be filled with N and A all the remaining three blanks can be filled with the remaining 3 letters in 3! ways=3*2*1=6
Ans:6 - 4 years agoHelpfull: Yes(6) No(0)
- N and A fixed
OKI - 3!=3*2*1= 6 - 4 years agoHelpfull: Yes(1) No(0)
- Nokia is a 5 letter word
N and A are at same position they can be arranged in 2! Ways .
And remaining 3 letters arranged in 3! Ways
So that, solution is 2!*3!=12 - 4 years agoHelpfull: Yes(0) No(4)
- 3!=1*2*3=6
- 4 years agoHelpfull: Yes(0) No(0)
- 4*3*2*1=24
- 4 years agoHelpfull: Yes(0) No(2)
- There are five letters in the given word.
Consider 5 blanks
The first blank and last blank must be filled with N and A all the remaining three blanks can be filled with the remaining 3 letters in 3! ways.
The number of words = 3! = 6. - 4 years agoHelpfull: Yes(0) No(0)
- N and A are fixed
So we have three letter i.e,=3!
so 31*2!=6 - 3 years agoHelpfull: Yes(0) No(0)
Hexaware Other Question