Mathematics, 17.09.2019 19:00 live4dramaoy0yf9

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

Answers: 1

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Answers

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

80%

step-by-step explanation:

Answer from: Quest

Ican't do any of the other stuff, but i can get you started. the answer, i believe, is $2.50.

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

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