C₇,₆ = 7

C₇,₁ = 7

C₇,₅ = 21

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A fórmula para o cálculo de uma combinação de n valores combinados p a p é dada por

[tex]\Luge{\boxed{\mathbf{C_{n,p}=\dfrac{n!}{p! \cdot(n-p)!}}}[/tex]

No nosso caso

C₇,₆

n = 7

p = 6

[tex]C_{n,p}=\dfrac{n!}{p! \cdot(n-p)!}\\\\\\C_{7,6}=\dfrac{7!}{6! \cdot(7-6)!}\\\\\\C_{7,6}=\dfrac{7!}{6! \cdot 1!}\\\\\\C_{7,6}=\dfrac{7!}{6! \cdot 1}\\\\\\C_{7,6}=\dfrac{7!}{6! }\\\\\\C_{7,6}=\dfrac{7\cdot 6!}{6!}\\\\\\Cancela\:\:o\:\:6!\\\\\\\mathbf{C_{7,6}=7}[/tex]

C₇,₁

n = 7

p = 1

[tex]C_{n,p}=\dfrac{n!}{p! \cdot(n-p)!}\\\\\\C_{7,1}=\dfrac{7!}{1! \cdot(7-1)!}\\\\\\C_{7,1}=\dfrac{7!}{1 \cdot 6!}\\\\\\C_{7,1}=\dfrac{7!}{6! }\\\\\\C_{7,1}=\dfrac{7\cdot 6!}{6!}\\\\\\Cancela\:\:o\:\:6!\\\\\\\mathbf{C_{7,1}=7}[/tex]

C₇,₅

n = 7

p = 5

[tex]C_{n,p}=\dfrac{n!}{p! \cdot(n-p)!}\\\\\\C_{7,5}=\dfrac{7!}{5! \cdot(7-5)!}\\\\\\C_{7,5}=\dfrac{7!}{5! \cdot 2!}\\\\\\\C_{7,5}=\dfrac{7!}{5! \cdot 2}\\\\\\C_{7,1}=\dfrac{7\cdot 6 \cdot 5!}{5! \cdot 2 }\\\\\\\\Cancela\:\:o\:\:5!\\\\\\C_{7,5}=\dfrac{7\cdot 6}{2 }\\\\\\C_{7,5}=\dfrac{42}{2 }\\\\\\\mathbf{C_{7,5}=21}[/tex]

No triângulo de Pascal:

1

1   1

1   2    1

1   3    3    1

1   4    6    4    1

1   5   10   10   5  1

1   6   15   20  15   6   1

1   7   21   35  35  21  7    1