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  • In how many ways 5-digit numbers can be formed using the digit 1,2,3,4,5,7,8,9 without repetition? How many of these are even numbers?Option: 1</strong

In how many ways 5-digit numbers can be formed using the digit 1,2,3,4,5,7,8,9 without repetition? How many of these are even numbers?

Option: 1

1080


Option: 2

2040


Option: 3

3060

 


Option: 4

4080


Answers (1)

best_answer

Using the permutation formula,

{ }^n P_r=\frac{n !}{(n-r) !}

Here,

\begin{aligned} & { }^8 P_5=\frac{8 !}{(8-5) !} \\ & { }^8 P_5=\frac{8 !}{3 !} \\ & { }^8 P_5=6720 \end{aligned}

Thus,

Therefore, in 6720 ways the 5-digits numbers can be formed using the digit 1,2,3,4,5,7,8,9.

To form an even number, the unit digit must be 2, 4, 8, or the fourth digit (thousands place) must be 2, 4, 8. For the remaining three digits, we can choose from the remaining six available digits.

Step 1: Count the number of choices for the units digit.

There are 3 choices for the units digit: 2, 4, or 8.

Step 2: Count the number of choices for the fourth digit (thousands place).

There are 3 choices for the fourth digit: 2, 4, or 8.

Step 3: Count the remaining choices for the first, second, and third digits.

After selecting the units digit and the fourth digit, there are 6 remaining digits to choose from for the first digit, 5 remaining digits for the second digit, and 4 remaining digits for the third digit.

Step 4: Multiply the choices from each step.

To calculate the total number of even 5-digit numbers without repetition, we multiply the number of choices from each step which is given by,

3(\text{ choices for the units digit }) \times 3(\text{ choices for the fourth digit }) \times 6(\text{ choices for the first digit }) \times 5 (\text{ choices for the second digit}) \times 4 (\text{choices for the third digit}) .

Thus,

3 \times 3 \times 6 \times 5 \times 4=1080

Therefore, there are 1,080 even 5-digit numbers that can be formed using the digits 1, 2, 3, 4, 5, 7, 8, and 9 without repetition.

 

Posted by

himanshu.meshram

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