drraju's problem please solve it

[munimM]

1. Given three numbers m, n and p. If L.C.M of ‘m’ and ‘n’ is ‘mn’ and L.C.M of ‘n’ and ‘p’ is ‘np’, then
Col A: L.C.M of ‘m’ and ‘p’
Col B: mp

2.A computer system digit code has 5 digits. In that ‘x’ repeats once, ‘y’ repeats twice and ‘z’ repeats twice. If the system accepts this code: xyzyz and yyxzz. Then
Col A: The number of combinations possible for the system code
Col B: xxx (some value)

[grefreak]

1..relationship cant be determined
2..30

[finix_regen]

2. 5!/2!*2!=30

[munimM]

1..relationship cant be determined
1..relationship cant be determined ..........why please explain with details clarification

[tushar4036]

1. if m=5 n=7 p=9
lcm of mn=35
lcm of np=63
lcm of mp=45 hence colA=colB

BUT m=6 n=7 p=8
lcm of mn=42
lcm of np=56
lcm of mp=24 which is less than 48

SO,the answer is D

[munimM]

thanks tusar

[alx1880]

A "for dummies" explanation for question #2 please.

[Curly213]

ohh ive read this question and ive understood it differently ...

2.A computer system digit code has 5 digits. In that ‘x’ repeats once, ‘y’ repeats twice and ‘z’ repeats twice. If the system accepts this code: xyzyz and yyxzz. Then
Col A: The number of combinations possible for the system code
Col B: xxx (some value)


i thought that ......If the system accepts this code: xyzyz and yyxzz... just accept this 2 codes ... so my calculation was 2C1*2C1*1C1 =4*2 =8....
but now iam gettin the real question thats just examples for the combination... to alex1880.... you have 5 digits,
so for the first choice u have 5 places
for the second u have 4 choices
and so on ....5!
now u have to divide by 2! because the letter y repeats twice (2)
and the same for the letter z
then u get this result 5!/(2!*2!) = 30

[alx1880]

Thanks Curly