Latest TCS Aptitude Question SOLUTION: Find the no. of all 6-digit natural no.s such that the sum of their digits is 10 and each of the digits 0, 1, 2, 3 occurs at least once in them. 1.490 2.386 3.520 4.620

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(#M40023721) TCS question Keep an EYE

Find the no. of all 6-digit natural no.s such that the sum of their digits is 10 and each of the digits 0, 1, 2, 3 occurs at least once in them.
1.490
2.386
3.520
4.620

Asked In TCS Karthika (10 years ago)

Solved Ritika M Read Solution (11)

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Ritika M  (10 years ago)

Answer is 490

The question says 0,1,2,3 should ATLEAST occur once in the digit, it does not say that they should be the only digits used

The combinations to get sum as 10 are:
0,1,2,3,2,2

So we get => 5*5*4*3*2*1/(3!) = 100 ways
(5 in the first place because zero cant be used there)

next- 0,1,2,3,3,1
So we get => 5*5*4*3*2*1/(2!. 2!) = 150

next- 0,1,2,3,4,0
So we get => 4*5*4*3*2*1/( 2!) = 240
(4 is multiplied for the first position because there are two zeroes)


100+150+240 = 490 !

Is this solution Helpfull? Yes (40) | No (6)  

Divya Srivastava  (10 years ago)

the correct answer is 490.
since the combination of digits having sum 10 are 1,2,3,0,2,2
or 1,2,3,0,3,1
or 1,2,3,0,4,0
finding the permutation for all three sets seperately we get (100+150+240) = 490

Is this solution Helpfull? Yes (16) | No (6)  

Karthika  (10 years ago)

Yes the correct choice from my buk was 490.
1.Three 2, One 3, One 1, One 0
(6! / 3!) - (5! / 3!)
= 5.5! / 3!
= 100 ways

2) One 2, Two 3, Two 1, One 0
(6! / 2!.2!) - (5! / 2!.2!)
= 5.5! / 4
= 150 ways

3) One 4, One 2, One 3, One 1, Two 0
(6!/2!) - 5!
= 240 ways

Total = 100 + 150 + 240
= 490 such numbers

Is this solution Helpfull? Yes (8) | No (1)  

Rudra  (10 years ago)

@karthika in the questions you didn't write 4, then why are u using 4 digit also???

Is this solution Helpfull? Yes (4) | No (1)  

Divya Srivastava  (10 years ago)

i m not saying that '0' occurs only at those places. it can occur at any place other than first place.
u can write the sets like 3,2,1,2,2,0
or 1,2,3,3,1,0
or 1,2,3,4,0,0
what matters is the choice of digits and its combination.

Is this solution Helpfull? Yes (3) | No   

Divya  (10 years ago)

DIVYA SRIVASTAVA ,the question is all about forming 6 digit numbers using four digits {0,1,2,3},but u used 4.howis it possible

Is this solution Helpfull? Yes (3) | No   

paramesh satya  (10 years ago)

Answer is 490

The question says 0,1,2,3 should ATLEAST occur once in the digit, it does not say that they should be the only digits used

The combinations to get sum as 10 are:
0,1,2,3,2,2

So we get => 5*5*4*3*2*1/(3!) = 100 ways
(5 in the first place because zero cant be used there)

next- 0,1,2,3,3,1
So we get => 5*5*4*3*2*1/(2!. 2!) = 150

next- 0,1,2,3,4,0
So we get => 4*5*4*3*2*1/( 2!) = 240
(4 is multiplied for the first position because there are two zeroes)


100+150+240 = 490 !

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

sujay suman  (10 years ago)

no are lyk=321040=360
=321031=180
=321022=120
which i greater than 620 so not sure is this r8 or wrong option??

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

dedeepya chowdary  (10 years ago)

as the sum is 10 & 1,2,3,0 are must so the remaning 4, can be as 2,2/3,1/4,0 & in a 6 digit if 0 comes frt it wll becm 5 digt so 386

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

shyamalabharathi  (10 years ago)

kartika can u xplain the solution once again clearly????

Is this solution Helpfull? Yes (1) | No   

Manish Kumar  (10 years ago)

how can u say that '0' occurs only at those places??

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