I : The number of five digit numbers that can be formed with $0, 1, 2, 3, 5$ which are divisible by $5$ is $18$ .
II : The sum of all four digit numbers that can be formed with $1, 2, 3, 4$ so that no digit being repeated in any number is $66660$ .
Which of the above statements is true?

Question

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Answer 1

Case I: $\Rightarrow$ last digit $= 0$
$C_{I} = 4! = 24$
Case II $\Rightarrow$ Last digit $= 5$
$C_{I I} = 3 \times 3! = 18$
$S = C_{I} + C_{I I} = 42$
II $\Rightarrow$ Sum here is
$S = 3! \times [1 + 2 + 3 + 4] \times [1 0^{0} + 1 0^{1} + 1 0^{2} + 1 0^{3}]$
$= 6 \times 10 \times [1111]$
$= 60 \times 1111$
$S = 66660$

Answer 2

Case I:

\Rightarrow

last digit

= 0

C_{I} = 4! = 24

Case II

\Rightarrow

Last digit

= 5

C_{I I} = 3 \times 3! = 18

S = C_{I} + C_{I I} = 42

II

\Rightarrow

Sum here is

S = 3! \times [1 + 2 + 3 + 4] \times [1 0^{0} + 1 0^{1} + 1 0^{2} + 1 0^{3}]

= 6 \times 10 \times [1111]

= 60 \times 1111

S = 66660