Respuesta :

Answer:  The required values are

[tex]^{11}C_8=165,~~~^7P_4=840.[/tex]

Step-by-step explanation:  We are given to evaluate the following combination and permutation :

[tex]h=^{11}C_8,~~~~~~k=^7P_4.[/tex]

We know that

[tex]^nC_r=\dfrac{n!}{r!(n-r)!},\\\\\\^nP_r=\dfrac{n!}{(n-r)!}.[/tex]

Therefore, we get

[tex]h=^{11}C_8=\dfrac{11!}{8!(11-8)!}=\dfrac{11\times10\times9\times8!}{*!\times3\times2\times1}=165[/tex]

and

[tex]^7P_4=\dfrac{7!}{(7-4)!}=\dfrac{7\times6\times5\times4\times3!}{3!}=840.[/tex]

Thus, the required values are

[tex]^{11}C_8=165,~~~^7P_4=840.[/tex]

BladeRunner212

165 and 840.

Further explanation

A combination is used to calculate how many ways to choose or know the various arrangements by not considering the order.

The formula for finding the number of different ways or number of combinations of n different objects taken r at the time is  

[tex]\boxed{\boxed{ \ _nC_r \ or \ C(n, r) = \frac{n!}{r!(n - r)!} \ }}[/tex]  

Given:

  • n = 11
  • r = 8

Question:

₁₁C₈ = ?

The Process:

Let us evaluate the value of ₁₁C₈.  

[tex]\boxed{ \ _{11}C_8 \ or \ C(11, 8) = \frac{11!}{8! \cdot (11 - 8)!} \ }[/tex]  

[tex]\boxed{ \ _{11}C_8 = \frac{11!}{8! \cdot 3!} \ }[/tex]  

Recall [tex]\boxed{n! = n \times (n-1) \times (n-2) \times ... \times 3 \times 2 \times 1}[/tex] as n factorial.  

[tex]\boxed{ \ _{11}C_8 = \frac{11 \times 10 \times 9 \times 8!}{8! \cdot 3!} \ }[/tex]  

We expand 11! because there are 8! inside it. Then we easily cross out 8! in the numerator and denominator.  

[tex]\boxed{ \ _{11}C_8 = \frac{11 \times 10 \times 9}{3 \times 2} \ }[/tex]

[tex]\boxed{ \ _{11}C_8 = 11 \times 5 \times 3 \ }[/tex]  

[tex]\boxed{ \ _{11}C_8 \ or \ C(11, 8) = 165 \ }[/tex]  

As a result, the expression ₁₁C₈ is 165.  

------------------------------------------------

A permutation is used to calculate how many ways to choose or know the various arrangements by considering the order.

The formula for finding the number of different ways or number of permutations of n different objects taken r at the time is

[tex]\boxed{\boxed{ \ _nP_r \ or \ P(n, r) = \frac{n!}{(n - r)!} \ }}[/tex]

Given:

  • n = 7
  • r = 4

Question:

₇P₄ = ?

The Process:

Let us evaluate the value of ₇P₄.

[tex]\boxed{ \ _{7}P_4 \ or \ P(7, 4) = \frac{7!}{(7 - 4)!} \ }[/tex]

[tex]\boxed{ \ _{7}P_4 = \frac{7!}{3!} \ }[/tex]

[tex]\boxed{ \ _{7}P_4 = \frac{7 \times 6 \times 5 \times 4 \times 3!}{3!} \ }[/tex]

We expand 7! because there are 3! Inside it. Then we easily cross out 3! in the numerator and denominator.

[tex]\boxed{ \ _{7}P_4 = 7 \times 6 \times 5 \times 4 \ }[/tex]

[tex]\boxed{ \ _{7}P_4 \ or \ P(7, 4) = 840 \ }[/tex]

As a result, the expression ₇P₄ is 840.

Learn more

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Keywords: evaluate  11C8 and 7P4, combination, permutation, the formula, 165, 840, n factorial

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