<s>[INST] How many five-digit numbers with distinct digits can be formed from the digits $0, 1, 2, 3, 4, 5, 6, 7$ under the given conditions? 
You can use the digits $0,1,2,3,4,5,6,7$.
You have to make a number of $5$ digits, above $9999$ (no $0$ in the start).
All digits should be different from each other, the number must have $3$ EVEN digits and $2$ ODD digits ($0$ is even), and the number should be divisible by $25$.

I need help in understanding the principle of one of the solving techniques which is:
letting everything roll as if $0$ can be in the start (let's call this I) - that time when $0$ is in the start (let's call this II):
What I don't understand is when do you use or don't use the other conditions ($3$ even and $2$ odd, divisible by $25$, ...). Is it in II and I or just one of them?
If someone can tell me what's wrong with this, I'll be glad:
Divide to $3$ cases: $25$, $50$, $75$.
first one is:
_ _ _ 2 5, (I)  >>>(pick 1 odd out of 3)*(pick 2 even out of 3)*3!
minus
0 _ _ 2 5, (II) >>>(pick 1 odd out of 3)*(pick 1 even out of 3)*2!
the second one is 50
_ _ _ 5 0 which is special >>> (pick 2 even out of 3)*(pick 1 odd out of 3)*3!
and the third one is
_ _ _ 7 5 >>>(pick 2 even out of 3)*3!
minus
0 _ _ 7 5, (II) >>>(pick 2 even out of 3)*2!
The answer in the book is $138$, and this doesn't add up to it. Thanks for your answers.
 [/INST] You have to take the conditions into account both when you count the five-digit strings and when you count the five-digit strings that begin with $0$.
The answer in the text is incorrect.
Method 1: We use your method.
Five-digit numbers of the form _ _ _ $25$ that use three even and two odd digits, all of which are distinct: Once we have used $2$ and $5$, we have three even and three odd digits remaining.  To form a five-digit string with three even digits and two odd digits, we must choose two of the three remaining even digits and one of the three remaining odd digits, then arrange the three selected numbers in the indicated positions, which can be done in
$$\binom{3}{2}\binom{3}{1}3! = 54$$
ways.
From these, we must subtract those five-digit strings of the form $0$ _ _ $25$ that use three even and two odd digits, all of which are distinct.  Once we have used $0$, $2$, and $5$, we have two even and three odd digits remaining.  We must choose one of the two remaining even digits and one of the three remaining odd digits, then arrange them in the indicated positions, which can be done in
$$\binom{2}{1}\binom{3}{1}2! = 12$$
ways.
Hence, we have 
$$\binom{3}{2}\binom{3}{1}3! - \binom{2}{1}\binom{3}{1}2! = 42$$
admissible numbers of the form _ _ _$25$.
Five-digit numbers of the form _ _ _$50$ that use three even and two odd digits, all of which are distinct:  Once we have used $5$ and $0$, we have three even and three odd digits remaining.  As shown above, there are 
$$\binom{3}{2}\binom{3}{1}3! = 54$$
strings of this form, all of which are admissible.
Five-digit numbers of the form _ _ _$75$ that use three even and two odd digits, all of which are distinct:  Once we have used $5$ and $7$, we have four even and two odd digits remaining.  We must use three of the four even digits in the remaining slots.  We must choose three of the four even digits, then arrange them in those slots, which can be done in 
$$\binom{4}{3}3! = 24$$
ways.
From these, we must subtract those strings of the form $0$_ _$75$ that use three even and two odd digits, all of which are distinct.  Once we have used $0$, $5$, and $7$, we are left with three even and two odd digits.  We must select two of the three even numbers and arrange them in the indicated positions, which can be done in
$$\binom{3}{2}2! = 6$$
ways.
Hence, there are
$$\binom{4}{3}3! - \binom{3}{2}2! = 18$$
admissible numbers of the form _ _ _$75$.
Total:  Since the three cases are mutually exclusive and exhaustive, the number of five-digit numbers with distinct digits which are divisible by $25$ composed from the set $\{0, 1, 2, 3, 4, 5, 6, 7\}$ is 
$$42 + 54 + 18 = 114$$
Method 2:  We do a direct count.
Five-digit numbers of the form _ _ _ $25$ that use three even and two odd digits, all of which are distinct:  We consider two cases, depending on whether the leading digit is even or odd.
Leading digit is even:  Since we cannot use $0$ or $2$ for the leading digit, the leading digit can be chosen in two ways from the two remaining even numbers.  We must use one of the two remaining even numbers (which include $0$ and the other unused even digit) and one of the remaining three odd numbers, then arrange them in the thousands and hundreds places, which can be done in 
$$\binom{2}{1}\binom{2}{1}\binom{3}{1}2! = 24$$
ways.
Leading digit is odd:  We must place one of the three remaining odd numbers in the first position.  We must choose two of the remaining even numbers for the remaining two slots, then arrange them in those slots, which can be done in
$$\binom{3}{1}\binom{3}{2}2! = 18$$
Hence, there are 
$$\binom{2}{1}\binom{2}{1}\binom{3}{1}2! + \binom{2}{1}\binom{2}{1}\binom{3}{1}2! = 42$$
admissible numbers of this form.
Five-digit numbers of the form _ _ _$50$ that use three even and two odd digits, all of which are distinct:  We consider two cases, depending on whether the leading digit is even or odd.
Leading digit is even:  Since we cannot use $0$ for the leading digit, we can choose the leading digit in three ways from the remaining even digits.  We must choose one even number from the remaining two even numbers and one odd number from the remaining three odd numbers, then arrange them in the remaining two positions.  There are
$$\binom{3}{1}\binom{2}{1}\binom{3}{1}2! = 36$$
such arrangements.
Leading digit is odd:  We can choose the leading digit in three ways from the remaining odd digits.  We must choose two of the remaining three even numbers then arrange them in the thousands and hundreds places.  There are
$$\binom{3}{1}\binom{3}{2}2! = 18$$
such arrangements.
Hence, there are
$$\binom{3}{1}\binom{2}{1}\binom{3}{1}2! + \binom{3}{1}\binom{3}{2}2!$$
admissible numbers of this form.
Five-digit numbers of the form _ _ _$75$ that use three even and two odd digits, all of which are distinct:  The three remaining digits must be even.  Since the  leading digit cannot be $0$, it can be chosen in three ways.  To fill the remaining two slots, we must choose two of the remaining three even numbers (which include $0$ and the other two unused even numbers), then arrange them in those slots, which can be done in
$$\binom{3}{1}\binom{3}{2}2! = 18$$
ways.
Total:  Since the three cases are mutually exclusive and exhaustive, the number of five-digit numbers with distinct digits which are divisible by $25$ composed from the set $\{0, 1, 2, 3, 4, 5, 6, 7\}$ is 
$$42 + 54 + 18 = 114$$
</s>