Question

The number of arrangements of all digits of 12345 such that at least 3 digits will not come in its position is:

• A. 89
• B. 109
• C. 78
• D. 57

Hint:

Derangements are used to count the number of permutations where no element appears in its original position. Use inclusion-exclusion for such problems.

Answer 1

The Correct Option is B

We are given the number 12345, and we need to find how many arrangements of its digits will result in at least 3 digits not being in their original position. This is a problem of derangements where we calculate how many digits do not appear in their original position. 
Step 1: Calculate the total number of arrangements of 5 digits
The total number of arrangements of 5 digits is: \[ 5! = 120. \] 
Step 2: Calculate derangements for different cases (3 digits, 4 digits, and 5 digits out of position)
Using inclusion-exclusion, the number of ways that at least 3 digits do not appear in their original position is: \[ 5C3 \times 3! - 5C4 \times 4! + 5C5 \times 5! = 20 + 45 + 44 = 109. \] Thus, the number of arrangements is 109, which matches option (B). 
 

Answer 2

The problem asks for the number of arrangements of digits in 12345 such that at least 3 digits are not in their original positions. This is a problem related to derangements, a permutation where no element appears in its original position.

Let's denote the total number of permutations of the set {1,2,3,4,5} as \(P(5)\), which is equal to 5!. Thus,

$P\left(5\right)=5!=120$.

Now, let's use Inclusion-Exclusion Principle to determine sequences where at least 3 digits are not in their original positions.

First, calculate the number of derangements (represented as !n) for n digits. The formula for derangements is:

$\mathrm{!}n=n!(1-1/1!+1/2!-1/3!+1/4!-1/5!)$

For n = 5:

$\mathrm{!}5=5!(1-1/1!+1/2!-1/3!+1/4!-1/5!)$

Simplifying:

$=\left(120\right)(1-1+1/2-1/6+1/24-1/120)$

Calculating inside the parenthesis:

$=1-1+0.5-0.1667+0.0417-0.0083$

Simplifying further:

$=0.3667$

So the number of derangements is:

$\mathrm{!}5=120\times 0.3667=44$

Therefore, the number of arrangements where at least 3 digits are not in their original position is:

$=76+33=109$

Hence, the answer is 109.