Question

How many combinations of two digit numbers having $8$ can be made from the following numbers?
$8,5,2,1,7,6$
A. $9$
B. $10$
C. $11$
D. $12$

Answer 1

The two digit number can have $8$ either in one’s place or in ten’s place and it can also have $8$ in both one’s and ten’s place. Excluding $8$ there are five other numbers with which it can form a two digit number.

Complete step-by-step answer:
Here given numbers are-
$8,5,2,1,7,6$
We have to find the number of combinations of two digit numbers having $8$made from these numbers.
Since it is not specified in the question whether $8$ has to be in one’s place or ten’s place so we can assume that the two digit number can have it in either place or even in both one’s and ten’s place.
Only one two digit number is possible with $8$ in both one’s and ten’s place=$88$
Now let us first find combinations of two digit number with $8$ on one’s place-
We have to find the digit in ten’s place because one’s place is fixed
Since there are five other numbers $5, 2, 1, 7, 6$ which can be placed in ten’s place, so the numbers are
$58, 28, 18, 78, 68$
These are $5$ combinations.
Now let us find combination of two digit number with $8$ in ten’s place-
We have to find the digit in one’s place because ten’s place is fixed.
Since there are five other numbers $5, 2, 1, 7, 6$ which can be placed in one’s place, so the numbers are
$85,82,81,87,86$
These are also five combinations.
So the total two digit numbers formed are $88,18,58,28,78,68,85,81,82,87,86$
So, total $11$ such two digit numbers can be formed.
Hence the correct option is ‘C’.

Note: We can also solve this question this way-
Since there are five other numbers except $8$ so total $10$ combinations can be made with $8$ in either ten’s place or one’s place. Only one combination is possible with $8$ in both one’s place and ten’s place. So total combinations are $11$.