To solve the problem of arranging the boys A, B, C, D, and E in four seats in increasing order of height such that C and E are never adjacent, we can break down the solution into clear steps.
### Step-by-Step Solution:
1. **Understanding the Problem**:
We have five boys: A, B, C, D, and E, with the height order A < B < C < D < E. We need to arrange four of them in increasing order while ensuring that C and E are not adjacent.
2. **Total Arrangements Without Restrictions**:
If we were to arrange any four boys from the five in increasing order, we can choose any four boys from the five. The number of ways to choose 4 boys from 5 is given by the combination formula:
\[
\binom{5}{4} = 5
\]
However, since they must be arranged in increasing order, there is only one way to arrange any selected group of four.
3. **Case Analysis**:
We will analyze two cases based on whether C and E are included in the selection.
**Case 1**: One of C or E is selected, but not both.
- If we select C, we can choose any three from A, B, D (since E cannot be selected). The only arrangement is A, B, C, D.
- If we select E, we can choose any three from A, B, C, D (since C cannot be selected). The only arrangement is A, B, D, E.
- Thus, for this case, we have:
\[
\text{Ways} = 1 \text{ (for C)} + 1 \text{ (for E)} = 2
\]
**Case 2**: Both C and E are selected.
- In this case, we must select two more boys from A, B, and D. The arrangements must still be in increasing order.
- If we select A and B, the arrangement is A, B, C, E.
- If we select A and D, the arrangement is A, C, D, E.
- If we select B and D, the arrangement is B, C, D, E.
- Thus, for this case, we have:
\[
\text{Ways} = 3 \text{ (A, B, C, E), (A, C, D, E), (B, C, D, E)} = 3
\]
4. **Total Arrangements**:
Now, we add the number of arrangements from both cases:
\[
\text{Total Ways} = 2 \text{ (from Case 1)} + 3 \text{ (from Case 2)} = 5
\]
5. **Final Answer**:
The total number of ways to arrange the boys in four seats in increasing order of height such that C and E are never adjacent is **5**.
