Latest TCS Aptitude Question SOLUTION: How many numbers can be formed using all of 1,2,3,4,5(without repetition),when the digit at the units place must be greater than that in the tenth place? (a)54(b)40(c)60(d)48(e)36

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(#M40001024) TCS Numerical Ability question Keep an EYE

How many numbers can be formed using all of 1,2,3,4,5(without repetition),when the digit at the units place must be greater than that in the tenth place?
(a)54(b)40(c)60(d)48(e)36

Solution

total methods without any restriction = 5*4*3*2*1 = 120
now for units place to be greater than tens place , only one method out of the set of two will be correct. ex: 12 & 21 only one will be valid so answer will be 120/2 = 60

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Asked In TCS Rini Mondal (13 years ago)

Solved Read Solution (6)

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Read Solution (6) :

Biswajit Das  (13 years ago)

when we take 1 as tenth place digit then in units place 2,3,4,5 can be used satisfying the condition.
so leftmost three digits(as it is non repetitive)can be combined into 3! way
total ways=4*3!
for 2=>>3*3!
for 3=>>2*3!
for4=>>1*3!
therefore total numbers will be=(4+3+2+1)*3!=60

option (c) is correct

Is this solution Helpfull? Yes (12) | No (1)  

Somil Nain  (13 years ago)

c) 60 taking separate cases
for 1 at tenth place 4 ways of filling unit place ,total=4*3*2*1=24
for 2 at tenth place 3 ways of filling unit place , total=3*3*2*1=18
for 3 at tenth place 2 ways of filling unit place , total=2*3*2*1=12
for 4 at tenth place 1 ways of filling unit place , total=1*3*2*1=6
for 1 at tenth place 0 ways of filling unit place
TOTAL=60

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Sourodip Ghosh  (13 years ago)

c)60
The nos. have to be 5 digit nos., since we are using all the five digits.
There are 10 ways to arrange nos. in ten's and unit's place to satisfy the condition in question ...
21,31,41,51,32,42,52,43,53,54
So now we have 10 possible combos for the last two digits: _ _ _ X X
We are left with the first 3.
There are 3 nos. left to fill these positions since 2 are getting used to fill up the last two positions. These 3 nos. can be arranged in these positions in 3P3 ways = 6 ways.
So final no. of combinations = 10*6 = 60

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

Tushar  (9 years ago)

For the condition to be satisfied, we need 2 nos out of 5, which can be chosen as 5C2 ways (gives 10), and the remaining three can be arranged between themselves which is 3! ways( gives 6).
Now multiply both
5C2 * 3! = 60.
Hope it helps..

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Tijo Jose  (13 years ago)

60

...1. --> 1*2*3*1*4 = 6*4
...2. --> 1*2*3*1*3 = 6*3
...3. --> 1*2*3*1*2 = 6*2
...4. --> 1*2*3*1*1 = 6*1
...5. --> 1*2*3*1*0 = 6*0

Total = 6*(0+1+2+3+4) = 60

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

Menda.haseena  (5 years ago)

60 is answer

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