Question

How many numbers greater than 56000 can be formed by using the digits 4, 5, 6, 7 and 8 with no digit repeated.

Answer 1

We solve this problem by assuming that there are 5 boxes in which 5 digits need to be placed in the certain order. Then we take all the possibilities of arranging the digits 4, 5, 6, 7 and 8 in each box of the 5 boxes such that each arrangement is greater than 56000.
First we find the number of numbers that are greater than 50000 then we find the number of numbers that are greater than 50000 and less than 56000 to get the required number of numbers.
We use the conditions that selecting $'r'$ objects from $'n'$ objects is given as ${}^{n}{{C}_{r}}$ where,
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$

Complete step-by-step answer:
We are given that there are 5 digits that are 4, 5, 6, 7 and 8.
Let us assume that there are 5 boxes in which the given 5 digits need to be arranged as follows

We are asked to find the number of numbers that are greater than 56000
Let us find the number of numbers that are greater than 50000
Now, let us take the possibilities of each box.
Here, we can see that the first digit should be greater than 5 because we are finding the numbers greater than 50000.
So, we can say that the possible digits in the first box are 5, 6, 7 and 8 which gives 4 possibilities.
Now, let us write the number of possibilities in first box then we get

We know that the conditions that selecting $'r'$ objects from $'n'$ objects is given as ${}^{n}{{C}_{r}}$
By using the above formula we get the number of ways of placing first digit as ${}^{4}{{C}_{1}}$
Now, let us take the second box.
Here as the number should be greater than 50000 we can say that all digits given that are 4, 5, 6, 7 and 8 can be placed but we already placed one digit in the first box.
So, we can say that the number of possibilities of second box is 4 that is

By using the selection formula we get the number of ways of placing digit in second box as ${}^{4}{{C}_{1}}$
Now, let us take the third box.
Here as the number should be greater than 50000 we can say that all digits given that are 4, 5, 6, 7 and 8 can be placed but we already placed two digits in first and second boxes.
So, we can say that the number of possibilities of second box is 3 that is

By using the selection formula we get the number of ways of placing digit in third box as ${}^{3}{{C}_{1}}$
Now, let us take the fourth box.
Here as the number should be greater than 50000 we can say that all digits given that are 4, 5, 6, 7 and 8 can be placed but we already placed three digits in first, second and third boxes.
So, we can say that the number of possibilities of third box is 2 that is

By using the selection formula we get the number of ways of placing digit in fourth box as ${}^{2}{{C}_{1}}$
Here, we can see that we have already placed 4 digits in the first four boxes.
So, we can say that the number of ways of placing the remaining digit in the last box can be done in 1 way.
Let us assume that the number of numbers that are greater than 50000 as ${{n}_{1}}$
We know that the number of ways of placing digits in each box is a permutation for the total number of ways. So, by using the permutations we get
$\Rightarrow {{n}_{1}}={}^{4}{{C}_{1}}\times {}^{4}{{C}_{1}}\times {}^{3}{{C}_{1}}\times {}^{2}{{C}_{1}}$
We know that the formula of combinations that is ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
By using the above formula we can say that ${}^{n}{{C}_{1}}=n$
By using this condition to above equation we get

& \Rightarrow {{n}_{1}}=4\times 4\times 3\times 2 \\\ & \Rightarrow {{n}_{1}}=96 \\\ \end{aligned}$$ Now, let us find the number of numbers that are greater than 50000 and less than 56000 Here, we can see that this is possible when the first two digits are 5 and 4 By placing 5 and 4 in first and second boxes respectively we get the number of possibilities as  Now, we can see that the remaining three boxes have 3 possibilities that are 6, 7 and 8 because we have already placed 5 and 4 in the first and second boxes. Let us assume that the number of numbers that are greater than 50000 and less than 56000 as $${{n}_{2}}$$ We know that the number of ways of distributing $$'n'$$ objects in $$'n'$$ places is given as $$n!$$ Now, by using the above formula we get $$\Rightarrow {{n}_{2}}=3!=6$$ Now, let us assume that the number of numbers that are greater than 56000 as $$N$$ then we get $$\begin{aligned} & \Rightarrow N={{n}_{1}}-{{n}_{2}} \\\ & \Rightarrow N=96-6 \\\ & \Rightarrow N=90 \\\ \end{aligned}$$ Therefore, we can conclude that there are a total of 90 numbers that are greater than 56000. **Note:** Students may make mistakes in taking the possibilities for second, third and fourth boxes. We have the number of possibilities in first box as Then we have the number of possibilities for second box as The number of possibilities for the second box is 4 because we have already placed one digit in the first box. As we are given that the numbers need not to be repeated the number of possibilities for second box should be 4 But, students may miss this point of not repeating the digits and take the number of possibilities of the second box as 5 which gives the wrong answer.