Latest TCS Aptitude Question SOLUTION: 2310 can be expressed as product of 3 factors in how many ways?

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2310 can be expressed as product of 3 factors in how many ways?

Asked In TCS MD (10 years ago)

Unsolved Read Solution (17)

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koduri tejaswini  (10 years ago)

2310=2*3*5*7*11
take each single factor and combinaton of ramaining factor for eg 11:factors are:11*6*35,
11*10*21,
11*14*15..
similarly for each factor we get 3 combinations
so total ways=3+3+3+3+3=15

Is this solution Helpfull? Yes (11) | No (5)  

Arun Kumar Parida  (10 years ago)

2310=2*3*5*7*11
no of ways= 5C3 + 5C2*3C2 + 1= 41(ans)

Is this solution Helpfull? Yes (10) | No (14)  

Prateek Pandey  (10 years ago)

Just to explain Arun's solution:
Factors are 2,3,5,7,11. We can group them in 2 ways: a group of 1,2,2 or a group of 1,1,3 factors
for the first group choose as 5c2 and out of left 3 as 3c2. So 5c2*3c2
for the sec group choose as 5c3 only (which is 5c3)
I am unable to understand why 1 is added at last

Is this solution Helpfull? Yes (8) | No (3)  

priyanka  (10 years ago)

ans is 10...as we have 2*3*5*7*11
and mke cobi to mke 3 factors so 5C3

Is this solution Helpfull? Yes (6) | No (12)  

MD  (10 years ago)

arun your answer is correct.Can u plzzzzzz explain how 5C2*3C2 and 1?
plz do tell the concept

Is this solution Helpfull? Yes (5) | No (2)  

Mahesh  (10 years ago)

The Prime factors of 2310 are 2, 3, 5, 7 and 11, as 2310 = 2*3*5*7*11.
Now, when 2310 is a product of three factors, two cases arise. (a) when 1 is one of the factors or the other case of two of the factors are 1's and (b) 1 is not one of the three factors.
Lets consider each of the cases individually with examples.

When two of the factors are 1's So, it should be 1 * 1* 2310. It can happen in only one way.
When 1 is one of the factors, for example 1*11*210 , here as one of the factors is 1, the remaining 5 prime factors are to be divided into two groups. So, five numbers into two groups, it can be 4 and 1 or 3 and 2
So, 1 * 4 factors * 1 factor - It can happen in 5 ways
Or, 1 * 3 factors * 2 factors, this can happen in 5c3 = 10 ways

Now the second case of 1 is not one of the factors,
5 factors are to be divided into three groups, this can be

3 factors * 1 factor * 1 factor- this can happen in 5c3 ways=10
2 factors * 2 factors & 1 factor- this can happen in 5c2 * 3c2 = 30 ways

Total = 1+5+10+10+30 = 56 ways

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

Atanu Roy  (10 years ago)

2310=2*3*5*11*7...so ans is 5c3+5c2*3c2(here you multiply 2 no and take other two)+1(for 210*1*1)=41

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

Anay Das  (10 years ago)

we cant take 5c3 because 2*3*1 or 5*3*7 and so on do not make 2310..
ans is 5C2*3C2+1(for 2310*1*1)=31
i think it is correct.

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

DHAKAD NEETENDRA SINGH  (10 years ago)

2310=2*3*5*7*11
combinatons of factors are
5C1*4C1*3C3=20
5C1*4C2*2C2=30
5C1*4C3*1C1=20
5C2*3C1*2C2=30
5C2*3C2*1C1=30
5C3*2C1*1C1=20
TOTAL no of ways = 20+30+20+30+30+20=150 ans...

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

lavanya  (10 years ago)

Ans:3^5=243

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

shefali  (10 years ago)

ans is 5c3 2310=2*11*5*3*7.
we have to find three distinct factors so we have to choose three out of 5 to make the combinations.

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Rituparna Bhattacharyya  (10 years ago)

ans: 15;
1/2 * 5C2 * 3c2

1/2 because we take 11,3 or 3,11 in combination..both result to same 33..

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Sriparna Boote  (10 years ago)

its 5C3
because we have 2310 = 2*3*5*7*11
and we have to express it as a product of 3 factors. so its 5C3.

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Utkarsha Singh  (10 years ago)

why 1 is added please explain...:?????????

Is this solution Helpfull? Yes | No (1)  

ravi teja  (10 years ago)

we have 2310 as 2*5*3*7*11
so now we should express as a product of 3 numbers then first lets take three numbers as
2*3*5 now wee have 2 number remaining to get again 2310 as its product
now this 2 numbers can be combined with 2 or 3 or 5 in 6 ways
other condition for selecting 3 number
this value of selection is 5c3 and now total is 6*5c3=60

Is this solution Helpfull? Yes | No (3)  

Siba Kumar Sahu  (10 years ago)

ans:
2310 = 2*3*5*7*11
here the factors of 2310 are different so we can write 2310 as product of
(any of 5 factor)*(any from rest 4 factors)*(product of rest factors)

this can be done in 5*4*1 = 20 ways

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jhinuk sikder  (10 years ago)

3 factors are 77*3*10

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