Pregunta
Evaluate the expression. a) \( { }_{11} C_{0} \) b) \( 11 P_{0} \) a) \( 11 C_{0}=\square \) (Simplify your answer.)
Ask by Hanson Savage. in the United States
Mar 12,2025
Solución de inteligencia artificial de Upstudy
Respuesta verificada por el tutor
Responder
a) \( 11C_{0} = 1 \)
b) \( 11P_{0} = 1 \)
Solución
**Step 1. Evaluate \( { }_{11}C_{0} \):**
The combination formula is given by
\[
{ }_{n}C_{r}=\frac{n!}{r!(n-r)!}.
\]
For \( n = 11 \) and \( r = 0 \),
\[
11C_{0}=\frac{11!}{0!(11-0)!}=\frac{11!}{0! \cdot 11!}.
\]
Since \( 0! = 1 \), this simplifies to
\[
11C_{0}=\frac{11!}{1 \cdot 11!}=1.
\]
**Step 2. Evaluate \( 11P_{0} \):**
The permutation formula is given by
\[
{ }_{n}P_{r}=\frac{n!}{(n-r)!}.
\]
For \( n = 11 \) and \( r = 0 \),
\[
11P_{0}=\frac{11!}{(11-0)!}=\frac{11!}{11!}=1.
\]
**Final Answers:**
a) \( 11C_{0} = 1 \)
b) \( 11P_{0} = 1 \)
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Simplifique esta solución Bonus Knowledge
The expression \( { }_{11} C_{0} \) represents the number of ways to choose 0 items from 11. The formula for combinations is given by \( {n \choose r} = \frac{n!}{r!(n-r)!} \). Plugging in our values, we get: \[ { }_{11} C_{0} = \frac{11!}{0!(11-0)!} = \frac{11!}{0! \cdot 11!} = 1 \] So, \( 11 C_{0} = 1 \). For \( 11 P_{0} \), which represents the number of ways to arrange 0 items from 11, we can use the formula for permutations: \( P(n, r) = \frac{n!}{(n-r)!} \). Applying it here, we get: \[ 11 P_{0} = \frac{11!}{(11-0)!} = \frac{11!}{11!} = 1 \] Thus, \( 11 P_{0} = 1 \). In summary: a) \( 11 C_{0} = 1 \) b) \( 11 P_{0} = 1 \)
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