Question

The English alphabets have 5 vowels and 21 consonants. How many words with two different vowels and two different consonants can be formed from the alphabet?

• A. \(2100 \times 2!\)
• B. \(210 \times 2!\)
• C. \(210 \times 4!\)
• D. \(2100 \times 4!\)

Hint:

When a question involves selecting letters and then arranging them, first use combinations to choose the required letters and then use permutations to arrange them. \[ \text{Total Ways} = \text{Selection Ways} \times \text{Arrangement Ways} \]

Answer 1

The Correct Option is D

Step 1: Select two different vowels.
There are \(5\) vowels in the English alphabet.
The number of ways to choose \(2\) different vowels is \[ {}^{5}C_{2} = \frac{5!}{2!\,3!} = 10 \]

Step 2: Select two different consonants.
There are \(21\) consonants in the English alphabet.
The number of ways to choose \(2\) different consonants is \[ {}^{21}C_{2} = \frac{21!}{2!\,19!} = 210 \]

Step 3: Form a four-letter word using the selected letters.
After choosing the letters, we have \[ 2 \text{ vowels } + 2 \text{ consonants } \] that is, a total of \(4\) distinct letters.
These \(4\) letters can be arranged in \[ 4! \] different ways.

Step 4: Apply the multiplication principle.
Therefore, the total number of required words is \[ {}^{5}C_{2} \times {}^{21}C_{2} \times 4! \] \[ = 10 \times 210 \times 4! \] \[ = 2100 \times 4! \]

Step 5: Final conclusion.
Hence, the number of words that can be formed is \[ \boxed{2100 \times 4!} \] Thus, the correct option is \[ \boxed{(4)\ 2100 \times 4!} \]