Answer
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Hint: There are four available digits out of which the code has to be formed. The largest possible number of trials means that the customer gets the code right in the last trial and all the previous trails fail. Thus, we have to find all the possible numbers of trials.
Complete step by step answer:
It is given in the question that the code has four digits 3, 5, 6 and 9.
To find the maximum number of possibilities to form the code, we will first consider the first space and number of possible numbers that can occupy that place.
So, the first place can occupy any one of 3, 5, 6 and 9 digits.
Thus, there are 4 possibilities for the first place.
Now, 3 digits are left for the 3 spaces. Therefore, second place can be occupied in 3 ways, since only three digits are available.
Now, 2 digits occupy the first and the second space, this means, 2 digits are left and there are 2 spaces left for these digits.
The third space can be occupied in 2 ways.
Now, when the third space is occupied in the third space, only one digit is left for the last space and there is only 1 way to place this digit.
Thus, first space : 4 ways, second space : 3 ways, third space : 2 ways and fourth space : 1 way.
Therefore, total number of ways to arrange 4 digits is (number of ways for 1st space $\times $ ways for 2nd space $\times $ ways for 3rd space $\times $ ways for 4th space)
$\Rightarrow $ (4 $\times $ 3 $\times $ 2 $\times $ 1) = 24 way.
Thus, the total number of trials the customer will require is 24.
Note: A short-cut method to find the number of possible ways to arrange n items in n spaces can be found out by n!, where n! is defined as n(n – 1)(n – 2)…..(1). Students are advised to go through concepts of permutation and combination to be able to solve such questions.
Complete step by step answer:
It is given in the question that the code has four digits 3, 5, 6 and 9.
To find the maximum number of possibilities to form the code, we will first consider the first space and number of possible numbers that can occupy that place.
So, the first place can occupy any one of 3, 5, 6 and 9 digits.
Thus, there are 4 possibilities for the first place.
Now, 3 digits are left for the 3 spaces. Therefore, second place can be occupied in 3 ways, since only three digits are available.
Now, 2 digits occupy the first and the second space, this means, 2 digits are left and there are 2 spaces left for these digits.
The third space can be occupied in 2 ways.
Now, when the third space is occupied in the third space, only one digit is left for the last space and there is only 1 way to place this digit.
Thus, first space : 4 ways, second space : 3 ways, third space : 2 ways and fourth space : 1 way.
Therefore, total number of ways to arrange 4 digits is (number of ways for 1st space $\times $ ways for 2nd space $\times $ ways for 3rd space $\times $ ways for 4th space)
$\Rightarrow $ (4 $\times $ 3 $\times $ 2 $\times $ 1) = 24 way.
Thus, the total number of trials the customer will require is 24.
Note: A short-cut method to find the number of possible ways to arrange n items in n spaces can be found out by n!, where n! is defined as n(n – 1)(n – 2)…..(1). Students are advised to go through concepts of permutation and combination to be able to solve such questions.
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