Question

The number of six-digit numbers that can be formed from the digits 1, 2, 3, 4, 5, 6 and 7 so that digits do not repeat, and the terminal digits are even is
A. 144
B. 72
C. 288
D. 720

Answer 1

Hint:Let us first fill the terminal digits of the number with even digits. After that fill all other places in a number. then find all the possible combinations and get the solution.

Complete step-by-step answer:
As we know that terminal digits in a number are only first and last digit.
So, let us fill the first and last place of the number with an even digit.
And we know that any digit is even if it is exactly divided by 2.
So, out of given digits there are only 3 even digits and that were {2, 4, 6}.
So, there are 3 possible digits that can be placed in the first place.
As we know that digits in a number are not repeated.
So, there will be 2 possible digits left for last place.
Now, we are left with 5 digits and 4 places. As we know that these four places can be filled with any number even or odd. But the digits cannot be repeated.
So, possible digits for second place will be 5.
Possible digits left for third place will be 4.
Possible digits left for fourth place will be 3.
And possible digits left for fifth place will be 2.
So, total number of numbers that can be formed with the digits {1, 2, 3, 4, 5, 6, 7} with no digits repeated and terminal digits as even will be 3$\times$5$\times$4$\times$3$\times$2$\times$2 = 720.
So, total possible numbers will be 720.
Hence, the correct option will be D.

Note: Whenever we come up with this type of problem then first, we will fill first and last place of the number with even digits and then other places. After placing a digit at any place, we subtract the number of digits left by 1. Because it is given that digits in a number cannot be repeated. And this will be the easiest and efficient way to find the solution of the problem.