Latest TCS Aptitude Question SOLUTION: In how many possible ways can write 3240 as a product of 3 positive integers a,b and c.

• Placement Papers
• /
• TCS
• /
• Numerical Ability
• /
• Probability

TCS Question Papers: TCS Interview Experience(90) TCS Company Profile

(#M40033219) TCS Numerical Ability question permutations and combinations Keep an EYE

In how many possible ways can write 3240 as a product of 3 positive integers a,b and c.

Asked In TCS sindhu (10 years ago)

Unsolved Read Solution (26)

Is this Puzzle helpful?   (68)   (53) Submit Your Solution Probability

Read Solution (26) :

jayesh  (10 years ago)

3240=2*2*2*5*3*3*3*3
so,
no. of ways=8!/(3!*4!)=420......

Is this solution Helpfull? Yes (39) | No (119)  

shireesha  (8 years ago)

Express 3240 as a factors of prime nos

3240 = 2 x 2 x 2 x 3 x 3 x 3 x 3 x 5 = 23 x 34 x 5

so these 3 nos of 2's can be distributed to 3 integres a,b,c in : [3+(3-1)]C (3-1) ways i.e 5C2 = 10 ways

Nos of ways in which 4 nos of 3's can be distributed to 3 integers : [4+(3-1)]C (3-1) ways i.e 6C2 = 15 ways

Nos of ways in which 1 nos of 5 can be distributed to 3 integers : [1+(3-1)]C (3-1) ways i.e 3C2 = 3 ways

So total ways = 10 x 15 x 3 = 450

Is this solution Helpfull? Yes (34) | No (3)  

Divya  (10 years ago)

3240 = 2^3*3^4*5... now x1+x2+x3 = 3 , so no. of solutions : 5C2.. similarily for 3 and 5, we get solns. as 6C2 and 3C2 respectively.. so ans : 5C2*6C2*3c2 = 450.

Is this solution Helpfull? Yes (24) | No (21)  

Himanshu Samtani  (9 years ago)

wrong answer 280 is coming by 8!/3!*4!.
please specify how did u get 420.

Is this solution Helpfull? Yes (17) | No (7)  

arun pandian  (9 years ago)

hey help me......say the ans how 8!/3!4!1!=420 but it comes 280 only ...........

Is this solution Helpfull? Yes (16) | No (7)  

Subhalakshmi.M  (8 years ago)

first take the L.C.M of 3240
2*2*2*5*3*3*3*3=280

by using permutation concept,we can express it as: 8!/[(3!)(1!)(4!)]
=1*2*3*4*5*6*7*8/[(1*2*3)(1)(1*2*3*4)]
=450

Is this solution Helpfull? Yes (11) | No (34)  

shailendra sethiya  (8 years ago)

3240==23×34×51=a×b×c
We have to distribute three 2's to a, b, c in 3+3−1C3−1=5C2=10 ways
We have to distribute four 3's to a, b, c in 3+4−1C3−1=6C2=15 ways
We have to distribute one 5 to a, b, c in 3 ways.
Total ways = 10×15×3=450 ways

Is this solution Helpfull? Yes (10) | No (4)  

Manikanta  (8 years ago)

2^3*3^4*5...
5c2 * 6c2 * 3c2=450

[a+b+c=n
n+r-1Cr-1 solutions]

Is this solution Helpfull? Yes (9) | No (1)  

krishnavamsy gonavaram  (9 years ago)

280
none of these

Is this solution Helpfull? Yes (7) | No (14)  

Abdul Rahman  (8 years ago)

guys how 2^3 and 3^4 and 5 is written as 5c2 6c2 3c2 plz explain

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

Buvaneshwari  (9 years ago)

it comes only 280..not 420..

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

sukanya sinha  (10 years ago)

8*5*9*9=3240
a=8,b=5,c=9(through l.c.m of 3240)

Is this solution Helpfull? Yes (4) | No (23)  

sindhu  (10 years ago)

this are the options given,a) 450 b)420 c)350 d)320

Is this solution Helpfull? Yes (4) | No (3)  

sujithra.M  (9 years ago)

3240= 2*2*2*5*3*3*3*3
8!/3!4!1!=420
so ans=420

Is this solution Helpfull? Yes (4) | No (14)  

Martin  (8 years ago)

The factors are 2^3*3^4*5,for 2^3 we get 5C2, for 3^4 we get 6C2, for 5 we get 5C2 and multiple all we get the answer as 450

Is this solution Helpfull? Yes (3) | No (7)  

Prashanth  (7 years ago)

Express 3240 as a factors of prime nos

3240 = 2 x 2 x 2 x 3 x 3 x 3 x 3 x 5 = 23 x 34 x 5

so these 3 nos of 2's can be distributed to 3 integres a,b,c in : [3+(3-1)]C (3-1) ways i.e 5C2 = 10 ways

Nos of ways in which 4 nos of 3's can be distributed to 3 integers : [4+(3-1)]C (3-1) ways i.e 6C2 = 15 ways

Nos of ways in which 1 nos of 5 can be distributed to 3 integers : [1+(3-1)]C (3-1) ways i.e 3C2 = 3 ways

So total ways = 10 x 15 x 3 = 450

Is this solution Helpfull? Yes (3) | No (2)  

Anup Vishwakarma  (4 years ago)

The factor 2 can be split among 3 divisors in 10 different ways:
0,0,3
0,1,2
0,2,1
0,3,0
1,0,2
1,1,1
1,2,0
2,0,1
2,1,0
3,0,0

The factor 3 can be split among 3 divisors in 15 different ways:
0,0,4
0,1,3
0,2,2
0,3,1
0,4,0
1,0,3
1,1,2
1,2,1
1,3,0
2,0,2
2,1,1
2,2,0
3,0,1
3,1,0
4,0,0

The factor 5 can be split among 3 divisors in 3 different ways:
0,0,1
0,1,0
1,0,0

Hence the total number of ways to write it as a product of 3 divisors is 10x15x3=450.

Is this solution Helpfull? Yes (3) | No   

Sandhya  (10 years ago)

ans:84
3240 can be written as:5*2*2*2*3*3*3*3

so we have to fill these numbers in multiples of
116(i.e. with 1num in 100's place & 1num in 10's place...)this combination is possible in 5 series 116---5
like this
125---10
134---17
224---20
233---19
242---13
total=84


this is in my view..
please specify if you know the options....

Is this solution Helpfull? Yes (2) | No (15)  

Pavithra  (6 years ago)

ans:450
(p+2)c2*(q+2)c2*(w+2)c2

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

sindhu  (10 years ago)

iam asking the possibility not the one way to get 3240.plz check it..once

Is this solution Helpfull? Yes | No   

sindhu  (9 years ago)

how 420 wll come as ans is 280????

Is this solution Helpfull? Yes | No   

raunaque.afroz  (9 years ago)

divya plz xpalin me y u hv tkn combinatn?

Is this solution Helpfull? Yes | No   

shivaram akumalla  (8 years ago)

Answer is 280 ..
2*2*2*3*3*3*3*5=3240.
8!/(3!*4!*1!)=280

Is this solution Helpfull? Yes | No (4)  

PRATAP BARE  (8 years ago)

3240=2*2*2*5*3*3*3*3
so,
no.of ways= 8!/3!*4! =8*7*6*5*4*3*2*1/3*2*4*3*2=280

Is this solution Helpfull? Yes | No (6)  

Divya Sharma  (5 years ago)

answer:
3240 = 23*34*51 = a*b*c

We have to distribute three 2's to a, b, c in 3+3−1C3−1 = 5C2 = 10 ways

We have to distribute four 3's to a, b, c in 3+4−1C3−1 = 6C2 = 15 ways

We have to distribute one 5 to a, b, c in 3 ways.

Total ways = 10*15*3 = 450 ways

Is this solution Helpfull? Yes | No   

kishore  (4 years ago)

2*2*2*5*3*3*3*3

Is this solution Helpfull? Yes | No (1)  

Please Login to Read Solution.