Math, asked by aarinsheik, 10 months ago


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

Answers

Answered by Anonymous
35

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.

Answered by Anonymous
6

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

Given ,

Repetition of letter is not allowed , so ,

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

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

 \starThe 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

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