There are 5 digits to choose from and 3 distinct digits must be chosen.

There are 5 possible digits to pick for the first digit, 4 possible digits to pick for the second digit (because we can't reuse the first digit) and 3 possible digits for the final digit (because we can't pick the first two digits).Therefore, there are 5⋅4⋅3=60 possible three digit numbers you can make with these digits without repeat. OR

In terms of Combinations, it would be; 5C1 x 4C1 x 3C1 = 60

Make them as fractions

72/100,120/100,224/100

so required l.c.m is (l.c.m of 72,120,224)/(h.c.f of 100,100,100)

=(63*160)/(100)

=100.8

Bcz we are making selections so we need Combinations and it is said we have to select at least 2, so It can be at least 2 or more than 2 i.e. till 7.

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P*4*13/100*12 -p*6*8/100*12=40 after solving p=12000

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take l.c.mof7,13,11 => 1001

Only 489489 is perfectly divisible by 1001.