Answer:
1. 56
2. n = 6
3. 1
Step-by-step explanation:
Formula:
\sf C(n,r) = \frac{n!}{r!(n-r)!}C(n,r)=
r!(n−r)!
n!
ㅤ
1. C(8, 3)
Solution:
\sf C(8,3) = \frac{8!}{3!(8-3)!}C(8,3)=
3!(8−3)!
8!
\sf C(8,3) = \frac{8!}{3!5!}C(8,3)=
3!5!
\sf C(8,3) = \frac{8 \times 7 \times 6 \times \cancel{5 \times 4 \times 3 \times 2 \times 1}}{(1 \times 2 \times 3) \times \cancel{(5 \times 4 \times 3 \times 2 \times 1)}}C(8,3)=
(1×2×3)×
(5×4×3×2×1)
8×7×6×
5×4×3×2×1
\sf C(8,3) = \frac{8 \times 7 \times 6}{1 \times 2 \times 3}C(8,3)=
1×2×3
8×7×6
\sf C(8,3) = \frac{8 \times 7 \times \cancel{6}}{\cancel{6}}C(8,3)=
6
8×7×
\sf C(8,3) = 8 \times 7C(8,3)=8×7
\sf C(8,3) = \boxed{\sf 56}C(8,3)=
56
2. C(n,4) = 15
\sf C(n,4) = 15C(n,4)=15
\sf \frac{n!}{4!(n-4)!} = 15
4!(n−4)!
=15
\sf \frac{n!}{(n-4)!} = 15 \times 4!
(n−4)!
=15×4!
\sf \frac{n!}{(n-4)!} = 360
=360
\sf \frac{n(n-1)(n-2)(n-3)\cancel{(n-4)!}}{\cancel{(n-4)!}} = 360
n(n−1)(n−2)(n−3)
\sf n(n-1)(n-2)(n-3) = 360n(n−1)(n−2)(n−3)=360
\sf n^4 - 6n^3 + 11n^2 - 6n = 360n
4
−6n
3
+11n
2
−6n=360
\sf n^4 - 6n^3 + 11n^2 - 6n - 360 = 0n
−6n−360=0
\sf (n+3)\big(\frac{ n^4 - 6n^3 + 11n^2 - 6n - 360 }{n+3}\big) = 0(n+3)(
n+3
n
−6n−360
)=0
\sf (n+3)(n^3-9n^2+38n-120) = 0(n+3)(n
−9n
+38n−120)=0
\sf (n+3)(n-6)\big(\frac{n^3-9n^2+38n-120}{n-6}\big) = 0(n+3)(n−6)(
n−6
+38n−120
\sf (n+3)(n-6)(n^2-3n+20) = 0(n+3)(n−6)(n
−3n+20)=0
\begin{gathered}\therefore \begin{cases} \sf n + 3 = 0 \implies n = -3\\\sf n - 6 = 0 \implies n = 6\\\sf n^2 - 3n + 20 = 0 \implies n = \frac{3\pm i \sqrt{71} }{2}\\\end{cases}\end{gathered}
∴
⎩
⎪
⎨
⎧
n+3=0⟹n=−3
n−6=0⟹n=6
−3n+20=0⟹n=
3±i
71
However, n ∈ N.
Since we have 3 solutions, and 6 is the only solution that is a natural number, we select 6 as our answer.
P.S. You can also try trial and error to solve for n.
3. C(9, 9)
\sf C(9,9) = \frac{9!}{9!(9-9)!}C(9,9)=
9!(9−9)!
9!
\sf C(9,9) = \frac{\cancel{9!}}{\cancel{9!}0!}C(9,9)=
0!
\sf C(9,9) = \frac{1}{0!} = \boxed{\sf 1}C(9,9)=
1
=
Planes
1. Plane HFKD
2. Plane GKJD
POINTS
1. I
2. F
3. G
4. K
5. J
6. D
7. H
LINES
1. Line IJ
2. Line DF
these are the answers, I hope it helpsss
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Answers & Comments
Answer:
Answer:
1. 56
2. n = 6
3. 1
Step-by-step explanation:
Formula:
\sf C(n,r) = \frac{n!}{r!(n-r)!}C(n,r)=
r!(n−r)!
n!
ㅤ
1. C(8, 3)
Solution:
\sf C(n,r) = \frac{n!}{r!(n-r)!}C(n,r)=
r!(n−r)!
n!
\sf C(8,3) = \frac{8!}{3!(8-3)!}C(8,3)=
3!(8−3)!
8!
\sf C(8,3) = \frac{8!}{3!5!}C(8,3)=
3!5!
8!
\sf C(8,3) = \frac{8 \times 7 \times 6 \times \cancel{5 \times 4 \times 3 \times 2 \times 1}}{(1 \times 2 \times 3) \times \cancel{(5 \times 4 \times 3 \times 2 \times 1)}}C(8,3)=
(1×2×3)×
(5×4×3×2×1)
8×7×6×
5×4×3×2×1
\sf C(8,3) = \frac{8 \times 7 \times 6}{1 \times 2 \times 3}C(8,3)=
1×2×3
8×7×6
\sf C(8,3) = \frac{8 \times 7 \times \cancel{6}}{\cancel{6}}C(8,3)=
6
8×7×
6
\sf C(8,3) = 8 \times 7C(8,3)=8×7
\sf C(8,3) = \boxed{\sf 56}C(8,3)=
56
ㅤ
2. C(n,4) = 15
Solution:
\sf C(n,4) = 15C(n,4)=15
\sf \frac{n!}{4!(n-4)!} = 15
4!(n−4)!
n!
=15
\sf \frac{n!}{(n-4)!} = 15 \times 4!
(n−4)!
n!
=15×4!
\sf \frac{n!}{(n-4)!} = 360
(n−4)!
n!
=360
\sf \frac{n(n-1)(n-2)(n-3)\cancel{(n-4)!}}{\cancel{(n-4)!}} = 360
(n−4)!
n(n−1)(n−2)(n−3)
(n−4)!
=360
\sf n(n-1)(n-2)(n-3) = 360n(n−1)(n−2)(n−3)=360
\sf n^4 - 6n^3 + 11n^2 - 6n = 360n
4
−6n
3
+11n
2
−6n=360
\sf n^4 - 6n^3 + 11n^2 - 6n - 360 = 0n
4
−6n
3
+11n
2
−6n−360=0
\sf (n+3)\big(\frac{ n^4 - 6n^3 + 11n^2 - 6n - 360 }{n+3}\big) = 0(n+3)(
n+3
n
4
−6n
3
+11n
2
−6n−360
)=0
\sf (n+3)(n^3-9n^2+38n-120) = 0(n+3)(n
3
−9n
2
+38n−120)=0
\sf (n+3)(n-6)\big(\frac{n^3-9n^2+38n-120}{n-6}\big) = 0(n+3)(n−6)(
n−6
n
3
−9n
2
+38n−120
)=0
\sf (n+3)(n-6)(n^2-3n+20) = 0(n+3)(n−6)(n
2
−3n+20)=0
\begin{gathered}\therefore \begin{cases} \sf n + 3 = 0 \implies n = -3\\\sf n - 6 = 0 \implies n = 6\\\sf n^2 - 3n + 20 = 0 \implies n = \frac{3\pm i \sqrt{71} }{2}\\\end{cases}\end{gathered}
∴
⎩
⎪
⎪
⎨
⎪
⎪
⎧
n+3=0⟹n=−3
n−6=0⟹n=6
n
2
−3n+20=0⟹n=
2
3±i
71
However, n ∈ N.
Since we have 3 solutions, and 6 is the only solution that is a natural number, we select 6 as our answer.
P.S. You can also try trial and error to solve for n.
ㅤ
3. C(9, 9)
Solution:
\sf C(n,r) = \frac{n!}{r!(n-r)!}C(n,r)=
r!(n−r)!
n!
\sf C(9,9) = \frac{9!}{9!(9-9)!}C(9,9)=
9!(9−9)!
9!
\sf C(9,9) = \frac{\cancel{9!}}{\cancel{9!}0!}C(9,9)=
9!
0!
9!
\sf C(9,9) = \frac{1}{0!} = \boxed{\sf 1}C(9,9)=
0!
1
=
1
Answer:
Planes
1. Plane HFKD
2. Plane GKJD
POINTS
1. I
2. F
3. G
4. K
5. J
6. D
7. H
LINES
1. Line IJ
2. Line DF
these are the answers, I hope it helpsss