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Logical Reasoning
Number Series
in how many ways can 2310 be expressed as a product of three factors
Read Solution (Total 9)
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- 2130=2*5*7*11*3*1
6 prime factors...to express this as a product of 3 we take, (n-1)c(r-1)
5c2=10 - 11 years agoHelpfull: Yes(16) No(15)
- ans is (41)
- 11 years agoHelpfull: Yes(6) No(1)
- 2310 = 2*3*5*7*11
We have to form a group of 3 out of these 5, it can be done in
5C2*3C2+5C3 ways i.e. 40 ways - 11 years agoHelpfull: Yes(4) No(9)
- three is right answer??
- 11 years agoHelpfull: Yes(3) No(4)
- Is The ans 60
- 11 years agoHelpfull: Yes(1) No(4)
- 2130 has 5 factor that are = 2,5,3,711;
so ways of making product of three factors= 5c3=5c2=10 - 9 years agoHelpfull: Yes(1) No(0)
- in 375 ways !!!
- 11 years agoHelpfull: Yes(0) No(0)
- 2310 = 2*3*11*5*7
therefore, x1+x2+x3+x4+x5 = 5
(5+3-1)C(3-1) = 21 - 9 years agoHelpfull: Yes(0) No(0)
- answer : 81.
2130 = 2*5*7*11*3
to experess this as a product of 3 factors let it be abc
2(only 1 digit & it it is to be counted 2 didits if it is 2^2) distributed to these 3 digits in [(1+3)C(3-1)]=3C2 ways = 3 ways
similary for 5,7,11,3 for each factor there will be 3 ways
hence total number of ways = 3*3*3*3*3= 243 ways.
Ans : 243 ways - 9 years agoHelpfull: Yes(0) No(0)
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