Mathematics, 17.09.2019 19:00 live4dramaoy0yf9

Evaluate
11c8 and 7p4.
11c8=
7p4=

Answers: 1

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Answer from: casrdunn7055

The required values are

$^{11}C_8=165,~~~^7P_4=840.$

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

$h=^{11}C_8,~~~~~~k=^7P_4.$

We know that

$^nC_r=\dfrac{n!}{r!(n-r)!},\\\\\\^nP_r=\dfrac{n!}{(n-r)!}.$

Therefore, we get

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

and

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

Thus, the required values are

$^{11}C_8=165,~~~^7P_4=840.$

Answer from: leo4687

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

$\boxed{\boxed{ \ _nC_r \ or \ C(n, r) = \frac{n!}{r!(n - r)!} \ }}$

Given:

n = 11r = 8

Question:

₁₁C₈ = ?

The Process:

Let us evaluate the value of ₁₁C₈.

$\boxed{ \ _{11}C_8 \ or \ C(11, 8) = \frac{11!}{8! \cdot (11 - 8)!} \ }$

$\boxed{ \ _{11}C_8 = \frac{11!}{8! \cdot 3!} \ }$

Recall $\boxed{n! = n \times (n-1) \times (n-2) \times ... \times 3 \times 2 \times 1}$ as n factorial.

$\boxed{ \ _{11}C_8 = \frac{11 \times 10 \times 9 \times 8!}{8! \cdot 3!} \ }$

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

$\boxed{ \ _{11}C_8 = \frac{11 \times 10 \times 9}{3 \times 2} \ }$

$\boxed{ \ _{11}C_8 = 11 \times 5 \times 3 \ }$

$\boxed{ \ _{11}C_8 \ or \ C(11, 8) = 165 \ }$

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

$\boxed{\boxed{ \ _nP_r \ or \ P(n, r) = \frac{n!}{(n - r)!} \ }}$

Given:

n = 7r = 4

Question:

₇P₄ = ?

The Process:

Let us evaluate the value of ₇P₄.

$\boxed{ \ _{7}P_4 \ or \ P(7, 4) = \frac{7!}{(7 - 4)!} \ }$

$\boxed{ \ _{7}P_4 = \frac{7!}{3!} \ }$

$\boxed{ \ _{7}P_4 = \frac{7 \times 6 \times 5 \times 4 \times 3!}{3!} \ }$

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

$\boxed{ \ _{7}P_4 = 7 \times 6 \times 5 \times 4 \ }$

$\boxed{ \ _{7}P_4 \ or \ P(7, 4) = 840 \ }$

As a result, the expression ₇P₄ is 840.

Learn moreLet us evaluate the value of ₁₀P₄ How many handshakes between four people? The most important element to scientists doing radiometric dating

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

Answer from: Quest

answer: need picture of problem

step-by-step explanation:

Answer from: Quest

Thats where they was ported and the fields that noone really knew much about needed tended, so they put slaves out there do free work and huge profits and so on the north decided to hold its own and was more different way of survival did not have same climate conditions to afford protecting slaves from winter so expenses be diff from south

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11c8 and 7p4.
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7p4=...

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