Latest TCS Aptitude Question SOLUTION: how many even natural number divisible by 5 can be formed using 0,1,2,3,4,5,6 with no repeatation of numbers

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(#M40030061) TCS question Keep an EYE

how many even natural number divisible by 5 can be formed using 0,1,2,3,4,5,6 with no repeatation of numbers

Asked In TCS pre (10 years ago)

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saurabh agarwal  (10 years ago)

6p1+6p2+6p3+6p4+6p5+6p6
=1956 Ans.

Is this solution Helpfull? Yes (25) | No (6)  

Sumana Sil  (10 years ago)

Hi! the correct answer is 1956.
Even number and divisible by five means: number has to end with 0.
Here we need to consider all the case. they are elucidated as follows :
1 digit number: none. ( as set of natural numbers doesn't include o)
2 digit number:place 0 at the unit's place, and we are left with only 6 options for the ten's digit. so 6.
3 digit numbers:--0. 6 options for hundred's place and 5 options for ten's place as repetitions are strictly forbidden.. so numbers possible :6*5=30. Similarly for the 4 digit , 5 digit, 6 digit and 7 digit numbers.
4 digit number:---0. 6*5*4=120.
5 digit numbers:----0. 6*5*4*3=360.
6 digit numbers:-----0. 6*5*4*3*2=720
7 digit numbers:------0. 6*5*4*3*2*1=720.
No other possibilities remain as repetitions are not allowed. Hope this has clarified your concern.
Regards.

Is this solution Helpfull? Yes (16) | No   

Anurag Kumar  (10 years ago)

1 digit no. =1(0)
2 digit no. =_0 =6
3 digit no. =_ _ 0 =6*5=30
4 digit no. =_ _ _ 0=6*5*4=120
5 digit no. =_ _ _ _ 0=6*5*4*3=360
6 digit no. =_ _ _ _ _ 0=6*5*4*3*2=720
7 digit no=_ _ _ _ _ _0=6*5*4*3*2*1=720
total=1957

Is this solution Helpfull? Yes (15) | No (6)  

SATYA NAYAK  (10 years ago)

6p1+6p2+6p3+6p4+6p5+6p6=1956
bcoz 2digit 3digit 4digit 5digit 6 digit and 7 digit numbers are considered

Is this solution Helpfull? Yes (3) | No   

urvashi chaudhuri  (10 years ago)

plz notice that it has asked for even natural nos. so we are not to consider nos ending with 5. we have to form nos ending with 0 only. so fix 0 at the unit's place and then the rest 6 digits in 6! ways. so its 720

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Ankit Gupta  (10 years ago)

bhai plz explain kaise hua mera 1236 a raha hai

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aditya kumar rauniyar  (10 years ago)

only no.s having last digit as 0 are both divisible by 5 and are even.
so,
case 1 : there are 0 no.s of 1 digits which is divisible by 5 and are even.
case 2 : there are 6 no.s of 2 digits (as tens place can be fill by 6 no.s only)
case 3 : there are 30 no.s of 3 digits (at hundreds place 6 no,tens place 5 no)
case 4 : there are 120 no.s of 4 digits (thous place=6 , h = 5 , t = 4 )
case 5 : there are 360 no.s of 5 digits (at ten thous place=6 ,t=5,h=4,t=3)
similarly,
case 6 : there are 720 no.s of 6 digits
case 7 : there are 720 no.s of 7 digits
hence total no.s are 0+6+30+120+360+720+1440 = 2676

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bhavna  (10 years ago)

answer will be 720

Is this solution Helpfull? Yes (1) | No (7)  

Pradeep sahu  (10 years ago)

is it 720?

Is this solution Helpfull? Yes (1) | No (5)  

SATYA NAYAK  (10 years ago)

6!=720 EASY HUH!!!!!

Is this solution Helpfull? Yes (1) | No (7)  

Shirshendu Ghosh  (10 years ago)

The no.s will be divisible by 5 only if last digit 0 or 5.
As no number can begin with 0 so, 6!/2!=360

Is this solution Helpfull? Yes (1) | No (3)  

kratika jain  (10 years ago)

1 digit no. =1 (i.e 0)
2 digit no. =_ 0 =5
3 digit no. =_ _ 0 =5*4=20
4 digit no. =_ _ _ 0=60
5 digit no. =_ _ _ _ 0= 120
6 digit no. = _ _ _ _ _ 0=120
total= 326

Is this solution Helpfull? Yes (1) | No (3)  

mehul  (10 years ago)

here 0 will be fixed at ones place
so 6! = 720

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Hrishabh Sobti  (10 years ago)

aditya kumar you added an extra 720.

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aditya kumar rauniyar  (10 years ago)

sorry i have by mistake added an extra 720 to my solution .....
hence the correct answer is 1956.

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Arnab Nandy  (10 years ago)

6!/2! is the answer

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Rohith Raju  (10 years ago)

the last place is filled with 0. then the remaining 6 places can be filled in 6! ways= 720

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santosh kumar senapati  (10 years ago)

6c1+6c2+6c3+6c4+6c5+6c6=2^6-1= 63 (no repetation of number)

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Aritra  (10 years ago)

7P2+7P6+7P5+7P4+7P3+7P2+7P1

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Pratibha Agrawal  (10 years ago)

1236
6+6*5+6*5*4+6*5*4*3+6*5*4*3*2

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DEEPANSHI SRIVASTAVA  (10 years ago)

6*5*4*3*2*1*1=720
even natural no. divisible by 5 must end with 0 i.e at unit place it will contain 0 value.

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Prabhat Yadav  (10 years ago)

i m confused plzzzz...suggest which is correct answer'

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ASHUTOSH KUMAR  (10 years ago)

urvashi@ is quetion me ye nai kahe raha ki only 7 seven digit no it may be 2 digit , 3 digit n 4 digit upto 7 digit possiable ho sakta hai so Saurabh agarwal is right

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Er. Anmol  (10 years ago)

Don't get confused between the formation of 7 digit numbers or even natural numbers. correct answer is 1956 for even numbers comprising of 2 digit no...upto 7 digit.
And 7 digit number divisible by 5 and even answer is 720

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amrtuha  (10 years ago)

case1: 1 digit=5=1 way
case2: 2 digit=_ 5;_ 0=>(5*1)+(6*1)=11 because the numbers ending with 5 or 0 are divisible by 5
case3: 3 digit=_ _ 0;_ _ 5=>(6*5*1)+(5*5*1)=55
case4: 4 digit=_ _ _ 0;_ _ _ 5=>(6*5*4*1)+(5*5*4*1)=220
case5: 5 digit=_ _ _ _ 0;_ _ _ _ 5=>(6*5*4*3*1)+(5*5*4*3*1)=660
case6: g digit=_ _ _ _ _ 0;_ _ _ _ _ 5=>(6*5*4*3*2*1)+(5*5*4*3*2*1)=1320
by adding all=>1320+660+220+55+11+1=2267.
i think this is the process... correct me if am wrong

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