Question

Consider all the 6 digit numbers that are formed by using the digits 1, 2, 3, 4, 5, and 6 where each digit is used only once. Each of such digit numbers has a property that for each digit, not more than 2 digits smaller than that digit appear to the right of that digit. Find the number of such 6 digit numbers having the desired property.

Answer 1

We solve this problem by using the permutations and combinations. We consider that there are 6 boxes representing the 6 places of 6 digit numbers. Then based on the given property of numbers we check the possible digits that can be placed in each box and then we fix one digit of possible digits in each box to get the required answer. The number of ways of selecting $'r'$ numbers from the set of $'n'$ numbers is given as ${}^{n}{{C}_{r}}$ where,
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$

Complete step-by-step solution:
We are asked to find the number of 6 digit numbers that have the given desired property.
Let us assume that there are 6 boxes that represent the 6 places of a 6 digit number.

We are given that the digits to be used are 1, 2, 3, 4, 5, and 6
We are given that the desired property of numbers is that for each digit, no more than 2 digits smaller than that digit appear to the right of that digit.
Here, we can see that we can place at most 2 digits that are less than a digit to the right of that digit.
Based on this property we can say that the digits that are possible to occupy first place are 1, 2, and 3 because if we place 4 in the first place then we get a possible number that has 1, 2, and 3 on the right of 4.
But we are given to place only up to 2 digits less than a digit to the right of that digit.
So, by writing the possibilities of the first place we get

Now, by using the same given property the digits that are possible in the second place are 1 to 4 because if we place digit 5 in second place then we get 3 digits that are less than 5 to the right of 5 which should not be considered.
By writing the possibilities in the second box we get

Similarly, by using the property to all the given boxes we get

Now, let us find the number of ways of placing digits in the possible ways we have.
We know that the number of ways of selecting $'r'$ numbers from the set of $'n'$ numbers is given as ${}^{n}{{C}_{r}}$ where,
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
By using the above formula we get the number of ways of placing 1 digit out of 3 digits in first place as

& \Rightarrow {}^{3}{{C}_{1}}=\dfrac{3!}{1!\left( 3-1 \right)!} \\\ & \Rightarrow {}^{3}{{C}_{1}}=3 \\\ \end{aligned}$$ We are given that we need to use one digit only once. So, let us assume that anyone digit of 1, 2, and 3 are placed in the first box. Then we can say that the number of digits that are remaining in second box is 3 So the number of ways of selecting 1 digit from 3 digits for second box is given as $$\begin{aligned} & \Rightarrow {}^{3}{{C}_{1}}=\dfrac{3!}{1!\left( 3-1 \right)!} \\\ & \Rightarrow {}^{3}{{C}_{1}}=3 \\\ \end{aligned}$$ Similarly, if two digits are placed in first and second boxes we have remaining 3 digits possible for the third box. So the number of ways of selecting 1 digit from 3 digits for the third box is given as $$\Rightarrow {}^{3}{{C}_{1}}=\dfrac{3!}{1!\left( 3-1 \right)!}=3$$ Now, if three digits are placed in the first, second, and third boxes we have the remaining 3 digits possible for the fourth box. So the number of ways of selecting 1 digit from 3 digits for the fourth box is given as $$\Rightarrow {}^{3}{{C}_{1}}=\dfrac{3!}{1!\left( 3-1 \right)!}=3$$ Here, we can see that we have placed 4 digits in the first four boxes. Now we have the remaining 2 digits. The number of ways of placing 2 digits in the remaining two places as $$\Rightarrow {}^{2}{{C}_{1}}=\dfrac{2!}{1!\left( 2-1 \right)!}=2$$ Let us assume that the total number of ways of arranging the numbers in given desire property as $$'N'$$ We know that the total number of ways that follow the given desired property is the permutations of individual possibilities we calculated above. By using the above condition we get the value of $$'N'$$ as $$\begin{aligned} & \Rightarrow N=3\times 3\times 3\times 3\times 2 \\\ & \Rightarrow N=162 \\\ \end{aligned}$$ Therefore, we can conclude that there are 162 numbers that satisfy the given desired property. **Note:** Students may make mistakes in calculating the possibilities of placing digits in second, third, and fourth boxes. The number of ways of placing digits in the first box is $$\begin{aligned} & \Rightarrow {}^{3}{{C}_{1}}=\dfrac{3!}{1!\left( 3-1 \right)!} \\\ & \Rightarrow {}^{3}{{C}_{1}}=3 \\\ \end{aligned}$$ We have a number of possible digits that can place in the second box is 4. We need to use a digit only once. Then the number of possible digits that can be placed in the second box is 3 as we placed one common digit already in the first place. So, we get the possible ways of placing a digit in the second box is $$\Rightarrow {}^{3}{{C}_{1}}=\dfrac{3!}{1!\left( 3-1 \right)!}=3$$ But, students may do mistake in the condition that using the digit only once and take the number of possible ways of placing a digit in the second box is $$\Rightarrow {}^{4}{{C}_{1}}=\dfrac{4!}{1!\left( 4-1 \right)!}=4$$