Question

The number of 6 digits numbers that can be formed from the digits 1,2,3,4,5,6 & 7 so that digits do not repeat and the terminal digits are even is:

Answer 1

Hint: The terminal digits of a number are the first and the last digits of a number. Find the number of ways these two digits can be filled by even digits. Also, find the numbers of ways in which the inner 4 digits can be filled with 5 options. Find the answer by multiplying the two.

Complete step by step answer:
The terminal digits of a number are the first and the last digits of the number. We have to fill these by even digits, and we have to choose from the digits 1,2,3,4,5,6 & 7.
Let us first consider the number of ways that the first digit of the number can be filled.
Since we can only consider even digits, the number of even digits in the sequence 1,2,3,4,5,6 & 7 are 3.
The even digits are 2,4 and 6. The first digit can be filled with any of these. Therefore we conclude that the number of ways to fill the first digit of the number is 3.
For the last digit of the number, we have to choose an even number from the sequence 1,2,3,4,5,6 & 7. There are 3 even digits present in the sequence, which are 2, 4 and 6.
But the repetition of digits in the number is not allowed. And one of these 3 digits is already being used to fill the first digit of the number.
Therefore, we can choose the last digit from the 3 even numbers but we have to exclude the number being used in the first digit of the number.
Therefore, only 2 options left to fill the last digit. Thus, we conclude that the number of ways to fill the last digit of the number is 2.
To fill the other 4 places left in the six digit number (except the first and last digit) , we have 5 options left, as 2 of them will be used to form the first and the last digit.
 We can find the number of ways that we can fill 4 places with 5 options by finding the value of \[^5{P_4}\].
$
  ^5{P_4} = \dfrac{{5!}}{{\left( {5 - 4} \right)!}} \\
   = \dfrac{{5!}}{1} \\
   = 5! \\
   = 120 \\
$

Thus, we have 120 ways of filling the middle 4 digits of the number.
Total number of ways to form the six digits number from the sequence 1,2,3,4,5,6 & 7 with terminal digits being even, and without repeat can be calculated by multiplying the choices to fill individual places.
Therefore, multiplying the above found choices, we get,
$
   = 3 \times 120 \times 2 \\
   = 720 \\
$
Thus the total number of ways is 720.

Note: The number of ways for filling $r$ places from the $n$options will be equal to $^n{P_r}$.
The formula for $^n{P_r}$ is $^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$. Total number of 6 digits can be calculated by multiplying the number of ways each digit can be filled.