Question

How many 4 – letter codes can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?

Answer 1

Hint:Fill 4 vacant places using 10 letters. As repetition is not allowed, the first place can be filled in 10 ways. The ${{2}^{nd}}$ place can be filled by the 9 remaining left and it goes on. Now multiply all the number of ways to get the required answer.

Complete step-by-step answer:
There are many ways of filling 4 vacant places in a 4 letter code. We have to fill it using the first 10 letters of the English alphabet. Thus, total number of letters = 10.
Now we have to fill the 4 vacant places in succession by 10 letters, keeping in mind that the repetition of letters is not allowed.
The first place can be filled in 10 different ways by any of the 10 letters. Now the second place can be filled by the remaining 9 letters in 9 ways. The third place can be filled with any of the remaining 8 letters in 8 different ways and the fourth place can be filled by any of the remaining 7 letters in 7 different ways.
Therefore, by multiplication principle the required number of ways in which 4 vacant places can be filled $=10\times 9\times 8\times 7=5040$.
Hence, 5040 four lettered codes can be formed using the first 10 letters of the English alphabet, if no letter is repeated.

Note: In the question It is said first 10 letters of the English alphabet. So, don’t confuse and take the 26 – digit alphabets. If repetition is allowed then the 4 – letter codes can be formed by, 4 vacant places can be filled = $=10\times 10\times 10\times 10=10000$ ways.