Question

I forgot my friend's 7 digit telephone number, but i remembered that the first two digits of the number are either 25 or 53. Also the number is even and the digit 2 appears once. What is the maximum number of trials i have to make before i can contact my friend?

Answer 1

Answer:

Consider two separate cose,

Case 1: 25_ _ _ _ _ (2 is already present)  

no. of possibilility = 9*9*9*9*4 (only four even numbers are possiblity in the last digit)

Case 2: 53_ _ _ _ _ Again here two different cases can occur

1. 2 is at the last digit

no. of possibilility = 9*9*9*9

2. 2 is not at the last digit

no. of possibilility = 4C1*9*9*9*4 (4C1 to choose any one place for 2, and 4 for remaing even numbers)

Add all the possibilities and get the answer.

Answer 2

Given:

• The number contains $7$ digits.
• The first two digits of the number are either $25$ or $53$.
• The number is even.
• The digit $2$ appears once.

To Find:

The maximum number of trials I have to make before I can contact my friend.

Solution:

From the given information it is clear that out of the $7$ digit positions, the first two positions are reserved for either $25$ or $53$.

Since the number is even, the last position can be filled by any of the digits $0, 2, 4, 6, 8$.

Now, consider two separate cases:

Case 1: The first two digits are $25$ i.e., $\[{\rm{2 5 \_ \_ \_ \_ \_}}\]$

Since $2$ has already appeared, the third, fourth, fifth, and sixth positions can be filled by any of the $9$ available digits $\[\left( {0,1,3,4,5,6,7,8,9} \right)\]$ and the last position can be filled by $0,4,6,8$.

Therefore, the total possible numbers in this case are $\[9 \times 9 \times 9 \times 9 \times 4 = 26244\]$.

Case 2: The first two digits are $53$ i.e., $\[{\rm{5 3 \_ \_ \_ \_ \_}}\]$

Here again, two cases arise.

1. Where $2$ is the last digit i.e, $\[{\rm{5 3 \_ \_ \_ \_ 2}}\]$

Since $2$ has already appeared, the third, fourth, fifth, and sixth positions can be filled by any of the $9$ available digits $\[\left( {0,1,3,4,5,6,7,8,9} \right)\]$.

Hence, the possible numbers are $\[9 \times 9 \times 9 \times 9 = 6561\]$.

2. Where $2$ is not the last digit $\[{\rm{5 3 \_ \_ \_ \_ \_}}\]$

$2$ can be filled in any of the four positions in $\[{}^4{{\rm{C}}_1} = 4\]$ ways. The remaining three positions can be filled by any of the $9$ available digits $\[\left( {0,1,3,4,5,6,7,8,9} \right)\]$ and the last position can be filled by $0,4,6,8$.

Hence, the possible numbers are $\[4 \times 9 \times 9 \times 9 \times 4 = 11664\]$.

Adding all the possible numbers from both cases, we get

The total possible numbers are:  $\[26244 + 6561 + 11664 = 44469\]$

Thus, the maximum number of trials I have to make before I can contact my friend is $44469$.

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