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In a hotel we can order two types of varieties, but we can make 6 more varieties in home. One can want the four varieties with two from hotel must. Find how many ways one can order.
Read Solution (Total 9)
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- for hotel 2/2=1
and home 6c2=15
ans=15 - 13 years agoHelpfull: Yes(25) No(2)
- out of four choices, 2 are from hotel so 2C2= 1
2 are from home, so 6C2=15
total choices = home and hotel
= 15*1= 15
- 13 years agoHelpfull: Yes(14) No(2)
- hotel 2c2 =1
house 6c2 =15
so ans is 1*15= 15 - 13 years agoHelpfull: Yes(3) No(2)
- As Q says 6 more verities means 2 of hotel & 6 of home total 8. But out of 4 dishes required 2 dishes are fixed (2C2). From remaining 6 dishes we have to choose 2 dishes (6C2). So Ans will be 2C2 * 6C2 = 1 * 15 = 15.
- 12 years agoHelpfull: Yes(3) No(0)
- There are 6 more varieties in home means total 8 varietis in home.....2 varieties from home....in question two from hotel must that means no selection for that , we have to select two varieties from home that is 8.So ans is 8c2=28...THIS ANS HUNDRED PERCENT CORRECT ...Guys follow this ...Ch.Raju
- 10 years agoHelpfull: Yes(2) No(2)
- ques is wrong it should be hotel in place of home
ans is 8C2*2= 56 - 13 years agoHelpfull: Yes(0) No(14)
- 6
4C2=(4*3)/(2*1)
=6 - 13 years agoHelpfull: Yes(0) No(15)
- please solve it nd tell me which is right answer
- 12 years agoHelpfull: Yes(0) No(0)
- here p(E)= 8c4=70,
2 must from hotel = 2c2=1,
remaining 2 from 6 =6c2=15,
total ways can one order is 15/70
ans is 15/70 - 9 years agoHelpfull: Yes(0) No(1)
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There are three frnds x, y , z gone to excursion with their girl frnds. there they wanted to find their weights but their GF’s are not accept to check their weight( all unnecessary data) . Then they check weights as x, y, z individually and then x and y, y and z, x and z ,then all(x,y,z). the last measure is 171. Then find the avg of all these seven measures.