Question: b) Each student is assigned an 8-character access code consisting of uppercase English letters (A–Z) and digits (0–9). Repetition is allowed.
i. How many different 8-character access codes can be formed? (03 Marks)
ii. From a group of 12 students, a committee of 4 students is to be selected for system testing. In how many ways can this be done? (03 Marks)
1. **Problem statement:**
We have two parts:
i. Calculate the number of different 8-character access codes formed from uppercase letters and digits with repetition allowed.
ii. Calculate the number of ways to select a committee of 4 students from 12.
2. **Part i: Number of 8-character access codes**
- The characters can be uppercase letters (A-Z) or digits (0-9).
- Number of possible characters = 26 (letters) + 10 (digits) = 36.
- Each of the 8 positions can be any of the 36 characters, repetition allowed.
**Formula:**
$$\text{Number of codes} = 36^8$$
3. **Calculation for part i:**
$$36^8 = 2821109907456$$
4. **Part ii: Number of ways to select 4 students from 12**
- This is a combination problem since order does not matter.
- Use the combination formula:
$$\binom{n}{r} = \frac{n!}{r!(n-r)!}$$
where $n=12$ and $r=4$.
5. **Calculation for part ii:**
$$\binom{12}{4} = \frac{12!}{4!\times 8!}$$
Simplify step-by-step:
$$= \frac{12 \times 11 \times 10 \times 9 \times \cancel{8!}}{4! \times \cancel{8!}} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1}$$
Calculate numerator and denominator:
$$= \frac{11880}{24} = 495$$
**Final answers:**
- i. Number of different 8-character access codes: $2821109907456$
- ii. Number of ways to select 4 students from 12: $495$