Question:
The number of arrangements of the letters of the word SEARCH such that no letter remains in its original position is:
The number of arrangements of the letters of the word SEARCH such that no letter remains in its original position is:
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Whenever a question asks for arrangements in which no object remains in its original position, immediately think of derangements. The standard value
\[
!6=265
\]
is frequently used in competitive examinations.
Updated On: Jun 10, 2026
- \(264\)
- \(265\)
- \(266\)
- \(267\)
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Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Concept:
The word SEARCH contains six distinct letters:
\[
S,\ E,\ A,\ R,\ C,\ H.
\]
The problem asks for the number of arrangements in which no letter occupies its original position.
Such arrangements are called
derangements. The number of derangements of \(n\) distinct objects is denoted by \[ !n. \] The formula is \[ !n = n!\left( 1-\frac1{1!} +\frac1{2!} -\frac1{3!} +\cdots +(-1)^n\frac1{n!} \right). \] Since SEARCH contains six distinct letters, we need \[ !6. \]
Step 1: Apply the derangement formula \[ !6 = 6! \left( 1-\frac1{1!} +\frac1{2!} -\frac1{3!} +\frac1{4!} -\frac1{5!} +\frac1{6!} \right). \] Since \[ 6!=720, \] we obtain \[ !6 = 720 \left( 1-1+\frac12-\frac16+\frac1{24}-\frac1{120}+\frac1{720} \right). \]
Step 2: Simplify the expression inside the bracket Taking LCM \(720\), \[ \frac{360-120+30-6+1}{720} = \frac{265}{720}. \] Thus, \[ !6 = 720\left(\frac{265}{720}\right). \] \[ !6 = 265. \] Therefore, \[ \boxed{265}. \]
Step 3: Verify using the standard derangement values The well-known derangement numbers are \[ !1=0, \] \[ !2=1, \] \[ !3=2, \] \[ !4=9, \] \[ !5=44, \] \[ !6=265. \] Hence our result is fully verified. Therefore, the required number of arrangements is \[ \boxed{265}. \]
derangements. The number of derangements of \(n\) distinct objects is denoted by \[ !n. \] The formula is \[ !n = n!\left( 1-\frac1{1!} +\frac1{2!} -\frac1{3!} +\cdots +(-1)^n\frac1{n!} \right). \] Since SEARCH contains six distinct letters, we need \[ !6. \]
Step 1: Apply the derangement formula \[ !6 = 6! \left( 1-\frac1{1!} +\frac1{2!} -\frac1{3!} +\frac1{4!} -\frac1{5!} +\frac1{6!} \right). \] Since \[ 6!=720, \] we obtain \[ !6 = 720 \left( 1-1+\frac12-\frac16+\frac1{24}-\frac1{120}+\frac1{720} \right). \]
Step 2: Simplify the expression inside the bracket Taking LCM \(720\), \[ \frac{360-120+30-6+1}{720} = \frac{265}{720}. \] Thus, \[ !6 = 720\left(\frac{265}{720}\right). \] \[ !6 = 265. \] Therefore, \[ \boxed{265}. \]
Step 3: Verify using the standard derangement values The well-known derangement numbers are \[ !1=0, \] \[ !2=1, \] \[ !3=2, \] \[ !4=9, \] \[ !5=44, \] \[ !6=265. \] Hence our result is fully verified. Therefore, the required number of arrangements is \[ \boxed{265}. \]
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