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Numerical Ability
Permutation and Combination
How many digits formed using 1,2,3,4 & 5. (If repitation is possible)
Read Solution (Total 15)
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- one digit-5
two digit-5^2
3 digit-5^3
4digit-5^4
5 digit-5^5..the total is addition of all those numbers - 11 years agoHelpfull: Yes(37) No(2)
- 5(SINGLE DIGIT)+25(TWO DIGIT)+125(THREE DIGIT)+625(FOUR DIGIT)+3125(FIVE DIGIT)=4025(ANSWER)
- 11 years agoHelpfull: Yes(10) No(5)
- 5^5
when repetition allowed =n^r - 11 years agoHelpfull: Yes(6) No(2)
- 5^5 digits=3125 can be formed
- 11 years agoHelpfull: Yes(3) No(1)
- As per the given question,first digit=5ways,second digit=5ways......fifth digit=5ways
so the answer is 5C1*5C1*5C1*5C1*5C1=5*5*5*5*5=5^5=3125. - 11 years agoHelpfull: Yes(3) No(0)
- 3245
repetition is possible means each number can come 5 times + all numbers together - 11 years agoHelpfull: Yes(2) No(0)
- each place can be filled in 5 ways thus its 5*5*5*5*5=5^5
- 11 years agoHelpfull: Yes(2) No(1)
- when repition is possible then no.of digits formed is 5^5
for suppose here they ask how many 4 digits can formed the the ans is 5 ^4.....
- 11 years agoHelpfull: Yes(1) No(0)
- Don't Know
- 11 years agoHelpfull: Yes(1) No(2)
- 5^5
bcoz in every position of the digit we can use that number again..
suppose a 5 digit number *****,,in 1st position I can use any number from 1,2,3,4,5,,so no. of choices is 5 to make a one digit no. ..similarly in 2nd position I can also use any number from 1,2,3,4,5 since repetition is allowed.. so to make a two digit total no of choices 5*5 .. to make five digit 5^5..uff!! i think this solution is helpful.. :) - 11 years agoHelpfull: Yes(1) No(0)
- 5^5 digits can be formed
- 11 years agoHelpfull: Yes(1) No(0)
- Ans: 5 digits are formed
Exp: in the question asked for digits not for numbers ....
Digit means containing only one digit - 11 years agoHelpfull: Yes(1) No(0)
- without repetition means simply n^n.
- 11 years agoHelpfull: Yes(1) No(0)
- n P r = n!/(n-r)! = 5 P 5 = 5!/0! =120
5 P 4 = 5!/1! = 120
5 P 3 = 5!/2! = 60
5 P 2 = 5!/3! = 20
5 P 1 = 5!/4! = 5
hence,
no. of digits formed = 5 P 5 + 5 P 4 + 5 P 3 + 5 P 2 + 5 P 1
= 120 + 120 + 60 + 20 + 5
= 325 - 11 years agoHelpfull: Yes(1) No(2)
- is solution for number of digits or number of numbers.
- 11 years agoHelpfull: Yes(0) No(2)
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