Determine the number of arrangements of letters of the word ALGORITHM if.
O is the first and T is the last letter.
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Determine the number of arrangements of letters of the word ALGORITHM if.
O is the first and T is the last letter.
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[tex]\Large\textrm{Pre-explanation}[/tex]
The word ALGORITHM consists of 9 alphabets.
[tex]\;[/tex]
The examples will be,
O_ _ _ _ _ _ _T
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We exclude the first and last letters. We have 7 alphabets to arrange again.
Lastly, we put O or T in the first and the last place.
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[tex]\Large\textrm{Explanation}[/tex]
We know,
[tex]\boxed{\rm^{n}P_{r}=\dfrac{n!}{(n-r)!}}[/tex]
If [tex]\rm n=r[/tex],
[tex]\boxed{\rm^{n}P_{n}=n!}[/tex]
[tex]\;[/tex]
We choose [tex]\rm 7[/tex] alphabets to arrange in order.
So we have,
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[tex]\Longrightarrow\ \rm^{7}P_{7}=\boxed{\rm7!\ ways}[/tex]
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So,
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[tex]\bf\Longrightarrow\ \boxed{\bf7!=5040\ ways}[/tex]
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[tex]\Large\textrm{Learn More}[/tex]
[tex]\textbf{- Formulas}[/tex]
Permutation
[tex]\boxed{\rm^{n}P_{r}=\dfrac{n!}{(n-r)!}}[/tex]
[tex]\;[/tex]
Combination
[tex]\boxed{\rm^{n}P_{r}=\dfrac{n!}{r!(n-r)!}}[/tex]
[tex]\;[/tex]
[tex]\rm^{n}P_{r}[/tex] where [tex]\rm n=r[/tex]
[tex]\rm\Longrightarrow\ ^{n}P_{n}=n![/tex]
[tex]\;[/tex]
[tex]\rm^{n}C_{r}[/tex] where [tex]\rm n=r[/tex]
[tex]\rm\Longrightarrow\ ^{n}C_{n}=1[/tex]
[tex]\;[/tex]
The factorial of [tex]\rm0[/tex]
[tex]\rm\Longrightarrow0!=1[/tex]
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[tex]\textbf{- Examples}[/tex]
Permutation:-
If we choose two students who will take president or vice-president, the number of ways to choose is a permutation. The order matters.
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Combination:-
If we choose two students who will clean up the class, the number of ways to choose is a combination. The order doesn't matter.