Question

The figures $4$, $5$, $6$, $7$ and $8$ are written in every possible order. The number of numbers greater than $56000$ is?
(A) $72$
(B) $96$
(C) $90$
(D) $98$

Answer 1

Hint: In the given question, we are provided with five digits $4$, $5$, $6$, $7$ and $8$. We have to find the number of ways such that the number formed by these digits is greater than $56000$. So, we will place the digits at each place value of the number one by one such that the number formed is greater than $56000$. We will use the multiplication rule of counting the total number of ways in which such numerals can be formed.

Complete answer:
So, we are to form numbers greater than $56000$ using the digits $4$, $5$, $6$, $7$ and $8$.
Now, the digit at the leftmost place cannot be $4$ at any cost as the number won’t be greater than $56000$. So, the number of numbers greater than $56000$ is the digit at the leftmost place is $4$ is zero.
Also, if the digit at the leftmost place is $5$, then it puts certain restrictions on the placement of the digits in the number. The digit to the immediate right of $5$ cannot be $4$ as the number would not be greater than $56000$ in this case. So, the digits allowed to the immediate right of $5$ are: $6$, $7$ and $8$. So, there are $3$ options for the digit at thousands place. After placing digit at thousands place value, the remaining places can be filled with remaining three digits including $4$. So, the remaining three digits can be arranged in $3! = 6$ ways.
So, the number of numbers greater than $56000$ is $3 \times 3! = 3 \times 6 = 18$ if the leftmost digit is $5$.
Now, if the digit at the leftmost place is $6$, $7$ or $8$. Then, there are no restrictions on the digits being placed at the remaining places as any number formed with the given set of digits and having $6$, $7$ or $8$ as the leftmost digit will be greater than $56000$ under any condition.
So, we have three options to fill the leftmost digit in this case. After filling the leftmost digit, the remaining four places can be filled with four digits in $4!$ ways.
So, the number of numbers greater than $56000$ is $3 \times 4! = 3 \times 24 = 72$ if the leftmost digit is $6$, $7$, or $8$.
Hence, we get the total number of numbers greater than $56000$ is $18 + 72 = 90$.
So, the option (C) is the correct answer.

Note:
In such a problem, one must know the multiplication rule of counting as consequent events are occurring. One must take care while doing the calculations and should recheck them so as to be sure of the final answer. We must find innovative methods of solving such questions as it opens our mind for various other such problems.