Question

A password consists of two alphabets from English followed by three numbers chosen from 0 to 3. If repetitions are allowed, the number of different passwords is:
1. ²⁶P₁ × ²⁶P₂ × ⁴P₁ × ³P₁ × ²P₁
2. (²⁶P₁)² × (⁴P₁)³
3. 26P1 × 26P2 × 4. (26P1 × 4P1)<span style="position: r

Answer 1

Correct Answer - Option 2 : (²⁶P₁)² × (⁴P₁)³

Concept:

Combinations: The number of ways in which r distinct objects can be selected simultaneously from a group of n distinct objects, is:

ⁿC_r= \(\rm \frac {n!}{r!(n-r)!}\).

Permutations: The number of ways in which r objects can be arranged in n places (without repetition) is:

ⁿP_r = \(\rm \frac{n!}{(n - r)!}\).

• nPr = r! × nCr.

  nP1 = nC1.

• n! = 1 × 2 × 3 × ... × n.
• 0! = 1.

 

Calculation:

There are a total of 5 symbols: 2 alphabets (out of 26) + 3 numbers (out of 4).

Since repetition is allowed, we will consider the number of arrangements of each symbol individually.

Each of the 2 alphabets can be arranged in 26P1 ways and each of the 3 numbers can be arranged in 4P1 ways.

Using the basic principle of counting, the total number of arrangements will be:

26P1 × 26P1 × 4P1 × 4P1 × 4P1

= (26P1)2 × (4P1)3.

 

Basic Principle of Counting:

If there are m ways for happening of an event A, and corresponding to each possibility there are n ways for happening of event B, then the total number of different possible ways for happening of events A and B are:

• Either event A alone OR event B alone: m + n.
• Both event A AND event B together: m × n.