Q.In how many ways can 5 Mathematics, 4 English and 3 Accountancy books can be arranged in a shelf if

(i) Same-subject books together: treat each subject as 1 block: $3!\times5!\times4!\times3!=103680$. …

(i) All books on the same subject together

1. Tie the 5 Maths books into one block, the 4 English into one block, the 3 Accountancy into one block — 3 blocks total.
2. Arrange the 3 blocks on the shelf: $3!=6$ ways.
3. Arrange the 5 distinct Maths books within their block: $5!=120$.
4. Arrange the 4 distinct English books within their block: $4!=24$.
5. Arrange the 3 distinct Accountancy books within their block: $3!=6$.
6. Total $=3!\times5!\times4!\times3! = 6\times120\times24\times6$.
7. Compute: $6\times120=720$; $720\times24=17280$; $17280\times6=103680$.

(ii) No two books on the same subject are adjacent

8. First arrange only the type-pattern of the 4 English (E) and 3 Accountancy (A) letters in a row of 7 slots — there are $\binom{7}{3}=35$ such letter-patterns in total (choosing which 3 of the 7 slots are A).
9. For a pattern broken into $r$ maximal "runs" of the same letter (e.g. EEE-A-E-AA has run-lengths $3,1,1,2$, so $r=4$ runs), the number of already-adjacent same-letter pairs is $k=7-r$ (a run of length $L$ contributes $L-1$ same-letter-adjacent pairs, summing to $7-r$ over the whole row).
10. Grouping the 35 E/A patterns by their run-count $r$ (equivalently by $k=7-r$) gives: $k=0$: 1 pattern, $k=1$: 6 patterns, $k=2$: 9 patterns, $k=3$: 12 patterns, $k=4$: 5 patterns, $k=5$: 2 patterns (these add to $1+6+9+12+5+2=35$ ✓). …