1. First, treat the books that must stay together- the books of one subject- as a single object. There are three such objects- "Math books", "Science books", "English books" so there are 3!= 6 ways to order the subjects. Since there are 3 Math books, there are 3!= 6 ways to order them within their group. Since there are there are 2!= 2 ways to order them. Since there are 4 English books, there are 4!= 24 ways to order them.

Putting that together, there are (6)(6)(2)(24)= 1728 ways to order all 9 books. That is considerably smaller than 7!= 5040 ways of ordering all 7 books without any restrictions.

2. Because the number must be odd, the "ones" digit must be odd: 3, 5, or 7.

Because the number must be less than 5000, the "thousands" digit must be one of 2, 3, or 4.

We are going to need to break this into "cases":

If the ones digit is 3, the thousands digit must be either 2 or 4. That leaves 5 digits from which to choose the remaining, middle, 3 digits: 5!/(5-3)!= 5!/2!= 60. That gives 2(60)= 120 possibilities for this case. (That "2" is for the two choices for thousands digit.)

If the ones digit is either 5 or 7, the thousands digit can be one of 2, 3, or 4. That leaves 5 digits from which to choose the remaining, middle digits. There are, as before, 60 ways to do that. That gives 2(3)(60)= 360 possibilities for this case. Putting those together, there are 120+ 360= 480 ways to do this.

3. A number is divisible by 5 if and only if its "ones" digit is 0 or 5. Taking either one of those leaves 9 digits for the first 6 digits. The leading digit cannot be 0 so that leaves 8 digits for the leading digit. That, finally, leaves 8 digits for the remaining, middle 5 digits: there are 8!/(8-5)!= 8!/2!= 20160 ways to do that. There are, then (2)(8)(20160)= 322560 ways to do this.