Question

4. How many 8-letter code can be formed using the 1st 10 letters of the English
alphabet, if no letter can be repeated?
​

Answer 1

Given:-

Number of letters = 10

Find:-

Number of alphabets that formed using first 10 letters of English alaphabet, without repeating.

Solution:-

Let us assume that code is in the form -

$\boxed{ \: \: \: \: }$ $\boxed{ \: \: \: \: }$ $\boxed{ \: \: \: \: }$ $\boxed{ \: \: \: \: }$ $\boxed{ \: \: \: \: }$ $\boxed{ \: \: \: \: }$ $\boxed{ \: \: \: \: }$ $\boxed{ \: \: \: \: }$

Number of 10 letters arranged in 1 box = 10

Number of 10 letters arranged in 2 box = 9

Number of 10 letters arranged in 3 box = 8

Number of 10 letters arranged in 4 box = 7

Number of 10 letters arranged in 5 box = 6

Number of 10 letters arranged in 6 box = 5

Number of 10 letters arranged in 7 box = 4

Number of 10 letters arranged in 8 box = 3

So,

Required number of code of 10 letters and 8 box = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3

=> 1814400

Hence, in 1814400 ways 8-letter code can formed.

Answer 2

$\large \underline{ \underline{ \sf \: Solution : \: \: \: }}$

Given ,

Repetition of letter is not allowed , so ,

$\star$The first place can be filled in 10 different ways by anyone of the first 10 letters of the english alphabet

$\star$The second place can be filled in 9 different ways by anyone of the remaining letters

$\star$The third place can be filled in 8 different ways by anyone of the remaining letters

$\star$Similarly , in 4th , 5th , 6th , 7th and 8th places can be filled in 7,6,5,4,3 different ways by anyone of the remaining letters

Therefore , by multiplication principle

$\to \sf No. \: of \: ways = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 \\ \\ \sf \to</p><p>No. \: of \: ways = 1814400$

Hence , the required number is 1814400