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Numerical Ability
Permutation and Combination
Q. How many 9 digit number can be formed from 1,2,3,4,5 which is divisible by 4,repetition is allowed?
Read Solution (Total 7)
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- digits are 1,2,3,4,5
for a no. to be divisible by 4, last two digit should be divisible by 4
last two digit may be 12,24,32,44,52 (5 ways)
last two places can be filled in 5 ways
so, remaining 7 places, each place can be filled in 5 ways => 5^7 ways
total no. formed = 5 * 5^7 = 5^8 - 10 years agoHelpfull: Yes(54) No(0)
- @rakesh ur genius
- 10 years agoHelpfull: Yes(15) No(2)
- rakesh bhai hats off
- 10 years agoHelpfull: Yes(2) No(2)
- Ans:
2*5^8
because 8th place can be filled with the digits 1 2 3 4 5 so this can also be filled in 5 ways but the 9th place can be filled in 2 ways because only 2 and 4 comes there hence ans will be 2*(5^8) - 10 years agoHelpfull: Yes(1) No(4)
- 5^7(3+2)=5^8
- 10 years agoHelpfull: Yes(0) No(1)
- as we consider last two digits of the no. to check the divisibility by 4.. so last two digits can be filled as:
12
24
32
44
52
so,here last digit can only be filled by 2 ways i.e. 2 & 4 while
second last digit can be filled by 5 ways i.e. 1,2,3,4,5
and other first 7 places can be filled in 5^7 ways as repeation is allowed.
so required result is: 5^7*5*2= 5^8*2. - 10 years agoHelpfull: Yes(0) No(0)
- For divisible by 4 we will see last 2 digits should be divisible by 4
we have 5 such no
so (5^7)* 5 - 8 years agoHelpfull: Yes(0) No(0)
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