
A variable name in a certain computer language must be either an alphabet or an alphabet followed by a decimal digit. The total number of different variable names that can exist in that language is equals to
( a ) 280
( b ) 390
( c ) 286
( d ) 296
Answer
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Hint: This question can be solved by finding the number of ways of choosing an alphabet out of 26 alphabets and adding it with the number of ways of choosing Alphabet followed by decimal value.
Complete step-by-step answer:
It is mentioned in the question that variable names in certain languages must be either an alphabet.
Now, total number of alphabet = 26.
Now, the number of ways of choosing an alphabet out of 26 alphabets = $ ^{26}{{C}_{1}} $
Also, the name of a variable in a certain computer language must be an alphabet followed by a decimal digit and also, it is given that the alphabet is followed by a decimal digit.
And, decimal digits can be 0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9.
So, the total number of decimal digits is equal to 10.
Then, the numbers of ways of choosing a decimal digit = $ ^{10}{{C}_{1}} $
And, the number of ways of choosing Alphabet followed by decimal value = $ ^{26}{{C}_{1}}{{\times }^{10}}{{C}_{1}}\times 1 $
The total number of different variable names that can exist in that language is equals to the number of ways of choosing an alphabet out of 26 alphabets and the number of ways of choosing Alphabet followed by decimal value together.
So, the total number of different variable names that can exist in that language is
= $ ^{26}{{C}_{1}} $ + $ ^{26}{{C}_{1}}{{\times }^{10}}{{C}_{1}}\times 1 $
$ ^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!} $ , where we choose r items from n items
= $ \dfrac{26!}{1!\left( 26-1 \right)!}+\dfrac{26!}{1!\left( 26-1 \right)!}\times \dfrac{10!}{1!\left( 10-1 \right)!}\times 1 $
= 26 + 26 $ \times $ 10
= 26 + 260
= 286
So, the total number of different variable names that can exist in that language is equals to 286.
So, the correct answer is “Option C”.
Note: While solving combination problems always remember for each parameter we need to find a number of ways individually. Always remember the formula of combination when we need to choose r items from total n items. Calculation should be accurate as expression can be lengthy and complex too. Also, here reverse order of decimal value and alphabet is not allowed, that is decimal value will occur only after alphabet only.
Complete step-by-step answer:
It is mentioned in the question that variable names in certain languages must be either an alphabet.
Now, total number of alphabet = 26.
Now, the number of ways of choosing an alphabet out of 26 alphabets = $ ^{26}{{C}_{1}} $
Also, the name of a variable in a certain computer language must be an alphabet followed by a decimal digit and also, it is given that the alphabet is followed by a decimal digit.
And, decimal digits can be 0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9.
So, the total number of decimal digits is equal to 10.
Then, the numbers of ways of choosing a decimal digit = $ ^{10}{{C}_{1}} $
And, the number of ways of choosing Alphabet followed by decimal value = $ ^{26}{{C}_{1}}{{\times }^{10}}{{C}_{1}}\times 1 $
The total number of different variable names that can exist in that language is equals to the number of ways of choosing an alphabet out of 26 alphabets and the number of ways of choosing Alphabet followed by decimal value together.
So, the total number of different variable names that can exist in that language is
= $ ^{26}{{C}_{1}} $ + $ ^{26}{{C}_{1}}{{\times }^{10}}{{C}_{1}}\times 1 $
$ ^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!} $ , where we choose r items from n items
= $ \dfrac{26!}{1!\left( 26-1 \right)!}+\dfrac{26!}{1!\left( 26-1 \right)!}\times \dfrac{10!}{1!\left( 10-1 \right)!}\times 1 $
= 26 + 26 $ \times $ 10
= 26 + 260
= 286
So, the total number of different variable names that can exist in that language is equals to 286.
So, the correct answer is “Option C”.
Note: While solving combination problems always remember for each parameter we need to find a number of ways individually. Always remember the formula of combination when we need to choose r items from total n items. Calculation should be accurate as expression can be lengthy and complex too. Also, here reverse order of decimal value and alphabet is not allowed, that is decimal value will occur only after alphabet only.
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