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  • How many 7-digit numbers containing the digits 1, 2, 4, 5, 6, 7, 8, and 9 can be created so that the digits 4 and 6 appear together?Option: 1 15640</

How many 7-digit numbers containing the digits 1, 2, 4, 5, 6, 7, 8, and 9 can be created so that the digits 4 and 6 appear together?

Option: 1

15640


Option: 2

17640


Option: 3

19440


Option: 4

16940


Answers (1)

best_answer

There are two ways to arrange the digits 4 and 5 next to each other: we can either have 45 or 54.

For each arrangement of the digits 4 and 5, we can treat them as a single digit and then arrange the remaining digits. 

Consider the numbers 4 and 5 to be single digits. This results in a set of 7 numbers:  1, 2, 45, 6, 7, 8, and 9.

The number can start with any of the 7 digits, so we have 7 choices for the first digit.

For the second digit, we again have 7 choices, since we can use any of the 7 remaining digits.

For the third digit, we have only 1 choice, which is the combined digit.

For the fourth digit, we have 6 choices, since we cannot use the digit that was used for the second digit.

For the fifth digit, we have 5 choices, since we cannot use either of the digits that were used for the second or fourth digit.

For the sixth digit, we have 4 choices, since we cannot use either of the digits that were used for the second, fourth, or fifth digit.

For the seventh digit, we have 3 choices, since we cannot use either of the digits that were used for the second, fourth, fifth, or sixth digit.

Thus, the total number of 5-digit numbers that can be formed using the digits 1, 2, 4, 5, 6, 7, 8, and 9 such that the digits 4 and 5 appear together in the number is given by,

7 \times 7 \times 1 \times 6 \times 5 \times 4 \times 3=17640

Therefore, the total number of ways the 7-digit number can be formed is 17640 ways.

Posted by

himanshu.meshram

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