Question

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

Answer 1

Hint: The numbers formed by the digits are given by the permutation of the digits with some conditions on them i.e numbers formed that are greater than 4000. For a number to be greater than 4000, thousands place can be filled by 3 numbers from given i.e 4,5 and 6 and remaining three digits can be taken by any of them then the total number of ways is the product of how each digit can be filled.

Complete step-by-step answer:
Now, we are given the digits 2, 3, 4, 5, and 6 which is a total of 5 digits.
We can form any number of digits from one to five.
We need to find the number of numbers which is greater than 4000. Hence, these numbers must be four-digit or five-digit only.
We first find the number of 4 digit numbers greater than 4000.
The four-digit has 4 digits and the first digit is the thousands place. The digits 2 and 3 are less than 4, hence they can't take the thousands place. While digit 4, 5, and 6 can take this place.
Now, we have the remaining 4 numbers and the remaining three digits can be taken by any of them to form a number greater than 4000.
Hence, the number of ways is given as follows:
${n_1} = 3 \times 4 \times 3 \times 2$
${n_1} = 72...............(1)$
Now, we find the number of 5 digit numbers greater than 4000. Observe that all five-digit numbers are greater than 4000. We have 5 digits and hence the number of five-digit numbers is given as follows:
${n_2} = 5 \times 4 \times 3 \times 2 \times 1$
${n_2} = 120..............(2)$
The total number of ways is the sum of the equation (1) and (2). Hence, we have:
$n = 72 + 120$
$n = 192$
Hence, the number of numbers that are greater than 4000 which can be formed using the digits 2, 3, 4, 5, 6 without repetition is 192.

Note: You can also use the permutation formula ${}^n{P_r} = \dfrac{{n!}}{{(n - r)!}}$ to find the total number of ways of arranging the digits to get the number greater than 4000.