IOQM P2 2024

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SomeonecoolLovesMaths

3131 posts

SomeonecoolLovesMaths

#1 h Sep 8, 2024, 9:25 AM

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The number of four-digit odd numbers having digits $1,2,3,4$ , each occuring exactly once, is:

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Archiiiiii

18 posts

Archiiiiii

#2 h Sep 8, 2024, 12:11 PM

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easy fpc problem. ans: 12

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Aaronjudgeisgoat

823 posts

Aaronjudgeisgoat

#3 h Sep 8, 2024, 1:10 PM

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Clearly, the last digit must be either $1$ or $3$ , giving us $2$ options for the last digit, and $3!=6$ options for the first $3$ . Therefore, there are a total of $2\cdot 6=\boxed{12}$ total numbers.

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megarnie

5531 posts

megarnie

#4 h Sep 8, 2024, 1:50 PM • 1 Y

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The number of positive four digit integers with $1,2,3,4$ and occurring exactly once is $4! = 24$ . But exactly half of these are odd (since the last digit is equally likely to be $1, 2, 3,$ or $4$ ), so the answer is $\boxed{12}$ .

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ItsAniBunny

13 posts

ItsAniBunny

#8 h Sep 12, 2024, 1:36 PM

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Classic combinatorics problem, answer is 12 cases. Their are two types of cases first is cases with first digit 1 and second is first digit is 3.

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SatisfiedMagma

450 posts

SatisfiedMagma

#9 h Sep 18, 2024, 5:50 AM

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For the entire number to be odd, fix the unit digit as $1,3$ . This can be done in $2$ ways. Rest of the digits can be permuted freely, particularly in $3!$ ways. So, the final answer is $\boxed{2 \cdot 3! = 12}$ . $\blacksquare$

This post has been edited 1 time. Last edited by SatisfiedMagma, Dec 5, 2024, 5:45 PM

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Sammy27

81 posts

Sammy27

#10 h Sep 25, 2024, 12:47 PM

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The last digit must either be $1$ or $3$ and so the rest of the digits can be selected in $3!$ ways in each case. Therefore the answer is $2\cdot 3!=\boxed{12}$ .

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Mathschampion

6 posts

Mathschampion

#11 h Oct 21, 2024, 1:11 PM

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The last digit is 1 or 3
And then these can be selected in 2*3! Ways
So , the answer is 12

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L13832

248 posts

L13832

#12 h Mar 7, 2025, 8:18 AM

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$2\cdot 3!=\boxed{12}$

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RachitSuckAtMath

5 posts

RachitSuckAtMath

#13 h Mar 11, 2025, 8:06 AM

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We have 2 cases -

Case-1 -> The number ends with 1

In this case, we fix 1 at the last position and the no. of ways to arrange the remaining 3 numbers is 3! = 6
So total numbers are 6

Case -2 -> The number ends with 3

Similar to Case-1, Total numbers in this case are also 6.

So total number of numbers is 6+6 =

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BackToSchool

1638 posts

BackToSchool

#14 h Yesterday at 12:46 PM

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megarnie wrote:

The number of positive four digit integers with $1,2,3,4$ and occurring exactly once is $4! = 24$ . But exactly half of these are odd (since the last digit is equally likely to be $1, 2, 3,$ or $4$ ), so the answer is $\boxed{12}$ .

Nice.

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