Question

Find the number of numbers that are greater than 4000 which can be formed using the digits 2, 3, 4, 5, 6 without repetition.

Answer 1

This problem deals with permutations and combinations. But here a simple concept is used. Although this problem deals with permutations only. Here factorial of any number is the product of that number and all the numbers less than that number till 1.
$\Rightarrow n! = n(n - 1)(n - 2).......1$
The number of permutations of $n$ objects taken $r$ at a time is determined by the formula which is used:
$\Rightarrow {}^n{p_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$

Complete step-by-step answer:
We have to find the no. of numbers greater than the given number 4000 using the given five digits.
So to find the numbers greater than 4000 can be a four-digit number or a five-digit number.
First finding the four-digit numbers which are greater than 4000.
There are four spots for the given five digits to fill in, as given below:

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So this 4 digit number has to be greater than 4000, so the first digit can’t be starting with the 2 or 3 from the given five digits. But the first digit should be starting with 4 or 5 or 6 from the given digits.
Now leaving the first blank, considering the next three blanks. The next three blanks have to be filled from the remaining four digits, as one of the digits from the given five digits.
If the first blank is occupied by 4, then the three blanks has to be filled with remaining four digits 2,3,5,6, which is done in the following no. of ways:
$\Rightarrow {}^4{p_3}$
Now if the first blank is occupied by 5, then the three blanks has to be filled with remaining four digits 2,3,4,6, which is done in the following no. of ways:
$\Rightarrow {}^4{p_3}$
Now if the first blank is occupied by 6, then the three blanks has to be filled with remaining four digits 2,3,4,5, which is done in the following no. of ways:
$\Rightarrow {}^4{p_3}$
So in total the no. of four digit numbers greater than 4000 from the digits 2,3,4,5,6 is given by:
$\Rightarrow {}^4{p_3} + {}^4{p_3} + {}^4{p_3}$
$\Rightarrow 3 \times {}^4{p_3} = 3 \times 24$
$\Rightarrow 3 \times {}^4{p_3} = 72$
$\therefore$ The no. of four digit numbers greater than 4000 from the digits 2,3,4,5,6 are 72.
Now the five-digit numbers that are greater than 4000 from the given five digits 2,3,4,5,6 is:
As the five-digit number should be greater than 4000 and has to be chosen from the given five digits.
So this can be arranged in :
$\Rightarrow 5! = 5 \times 4 \times 3 \times 2 \times 1$
$\Rightarrow 5! = 120$
So the five-digit numbers greater than 4000 formed by the given 5-digits are 120.
So the total no. of numbers greater than 4000 formed by the given digits 2,3,4,5,6 are given by:
$\Rightarrow 72 + 120$
$\Rightarrow 192$

Final Answer: The no. of numbers that are greater than 4000 which can be formed using the digits 2, 3, 4, 5, 6 without repetition are 192.

Note:
Here while solving such kind of problems if there is any word of $n$ letters and a letter is repeating for $r$ times in it, then it can be arranged in $\dfrac{{n!}}{{r!}}$ number of ways. If there are many letters repeating for a distinct number of times, such as a word of $n$ letters and ${r_1}$ repeated items, ${r_2}$ repeated items,……. ${r_k}$ repeated items, then it is arranged in $\dfrac{{n!}}{{{r_1}!{r_2}!......{r_k}!}}$ number of ways.