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The number of permutation of $\{1,2,3,4,5\}$ that keep at least one integer fixed is.

  1. $81$
  2. $76$
  3. $120$
  4. $60$
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5 Answers

Best answer
14 votes
14 votes

It is based on inclusion exclusion principle:

Total arrangements - derangements 

$\begin{align}&=5!-[5!-^5C_1(4!)+^5C_2(3!)-^5C_3(2!)+^5C_4(1!)-^5C_5(0!)]\\&=^5C_1(4!) - ^5C_2(3!) + ^5C_3(2!) -^5C_4(1!) +^5C_5(0!)\\&=120 - 60 + 20 - 5 + 1\\ &= 76.\end{align}$ 

Correct Answer: $B$

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1 votes

Here many people is mis understand this question so that they get 120 as answer which is wrong. So we first understand the question.

Given that ATLEAST ONE INTEGER FIXED, which means you can take any permutation and in that permutation atleast one integer is on its actual place and others are not.

EG:  12345 is given so possible permutation with one integer(1) is fixed 13245, 14523, 15432 etc... now one integer(2) is fixed 32154, 42153, 52134 etc...likewise only one integer is fixed

Now two integer(12) is fixed 12534, 12453 etc... likewise we have to fix 3 and 5 integer also, Here keep in mind that we can't make any permutation with fixing 4 integers coz in that case last integer has to fit in actual place coz no dearrangement is possible.   

Here all the permutations in which every integer is dearrange is not included in our answer.

so our answer is 5!-D5=76        where D5 is dearrangement of 5 integer.                                                               

1 votes
1 votes

Correct Option: (B) 76

Let, $S = \left \{ 1, 2, 3, 4, 5 \right \}$,

$S_1, S_2, S_3, S_4, S_5$ be the the five integers,

$|S_i|$ where $i \in S$ mean that integer $S_i$ is fixed .

Now, we want to find the no. of permutations in which at least 1 integer is fixed.

In other words, we want to find $|S_1 \cup  S_2 \cup S_3 \cup S_4 \cup S_4 \cup S_5|$.

Now, according to the Principle of Arrangement and De-arrangement, we have

 

$|S_1 \cup  S_2 \cup S_3 \cup S_4 \cup S_4 \cup S_5| = \sum_{i \in S}|S_i| - \sum_{i, j \in S \ \land \  i \neq j}|S_i \cap S_j| + \\ \sum_{i, j, k \in S \ \land \ i \neq j \neq k}|S_i \cap S_j \cap S_k| - \sum_{i, j, k, l \in S \ \land \ i \neq j \neq k \neq l}|S_i \cap S_j \cap S_k \cap S_l| + \\ \sum_{(i, j, k, l, m \in S \ \land \ i \neq j \neq k \neq l \neq m}|S_i \cap S_j \cap S_k \cap S_l \cap S_m|$

 

Now,

$\sum_{i \in S}|S_i| = \binom{5}{1} \times 4! = 120$

$\sum_{i, j \in S \ \land \  i \neq j}|S_i \cap S_j| =  \binom{5}{2} \times 3! = 60$

$\sum_{i, j, k \in S \ \land \ i \neq j \neq k}|S_i \cap S_j \cap S_k| = \binom{5}{3} \times 2! = 20$

$\sum_{i, j, k, l \in S \ \land \ i \neq j \neq k \neq l}|S_i \cap S_j \cap S_k \cap S_l| = \binom{5}{4} \times 1! = 5$

$\sum_{(i, j, k, l, m \in S \ \land \ i \neq j \neq k \neq l \neq m}|S_i \cap S_j \cap S_k \cap S_l \cap S_m| = \binom{5}{5} = 1$

 

$\therefore$ We finally have,

$|S_1 \cup  S_2 \cup S_3 \cup S_4 \cup S_4 \cup S_5| = 120 - 60 + 20 - 5 +1 = 76$

 

 

0 votes
0 votes

Derangement means no elements should appear in its original position so derangement of 5 means all 5 elements do not appear on its original position.

So if we subtract D5 from total ways we shall get numbers in which at least one element is fixed because we already subtracted those elements where all 5 elements do not appear on same position hence only that elements are remaining where at least one element is fixed in its original position.

Total ways = 5!
D5(derangement of 5) = 44 (apply formula)
 Desired output:     5!-44 = 76

Hence correct answer is option B

 

 

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