Math  /  Discrete

QuestionEvaluate the following: a) 5C3{ }_{5} C_{3} c) (64)\binom{6}{4} e) (126)\binom{12}{6} b) 9C8{ }_{9} C_{8} d) 10C0{ }_{10} C_{0} f) 8C1{ }_{8} C_{1}

Studdy Solution

STEP 1

1. The notation nCr{ }_{n} C_{r} or (nr)\binom{n}{r} represents the binomial coefficient, which calculates the number of combinations of n n items taken r r at a time.
2. The formula for the binomial coefficient is (nr)=n!r!(nr)!\binom{n}{r} = \frac{n!}{r!(n-r)!}, where n! n! denotes the factorial of n n .

STEP 2

1. Evaluate 5C3{ }_{5} C_{3}.
2. Evaluate (64)\binom{6}{4}.
3. Evaluate (126)\binom{12}{6}.
4. Evaluate 9C8{ }_{9} C_{8}.
5. Evaluate 10C0{ }_{10} C_{0}.
6. Evaluate 8C1{ }_{8} C_{1}.

STEP 3

Evaluate 5C3{ }_{5} C_{3} using the formula:
5C3=5!3!(53)!=5×4×3!3!×2×1=5×42×1=202=10{ }_{5} C_{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4 \times 3!}{3! \times 2 \times 1} = \frac{5 \times 4}{2 \times 1} = \frac{20}{2} = 10

STEP 4

Evaluate (64)\binom{6}{4} using the formula:
(64)=6!4!(64)!=6×5×4!4!×2×1=6×52×1=302=15\binom{6}{4} = \frac{6!}{4!(6-4)!} = \frac{6 \times 5 \times 4!}{4! \times 2 \times 1} = \frac{6 \times 5}{2 \times 1} = \frac{30}{2} = 15

STEP 5

Evaluate (126)\binom{12}{6} using the formula:
(126)=12!6!×6!=12×11×10×9×8×7×6!6!×6×5×4×3×2×1=665280720=924\binom{12}{6} = \frac{12!}{6! \times 6!} = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6!}{6! \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} = \frac{665280}{720} = 924

STEP 6

Evaluate 9C8{ }_{9} C_{8} using the formula:
9C8=9!8!(98)!=9×8!8!×1=9{ }_{9} C_{8} = \frac{9!}{8!(9-8)!} = \frac{9 \times 8!}{8! \times 1} = 9

STEP 7

Evaluate 10C0{ }_{10} C_{0} using the formula:
10C0=10!0!(100)!=10!1×10!=1{ }_{10} C_{0} = \frac{10!}{0!(10-0)!} = \frac{10!}{1 \times 10!} = 1

STEP 8

Evaluate 8C1{ }_{8} C_{1} using the formula:
8C1=8!1!(81)!=8×7!1×7!=8{ }_{8} C_{1} = \frac{8!}{1!(8-1)!} = \frac{8 \times 7!}{1 \times 7!} = 8
The evaluated values are: a) 1010 b) 99 c) 1515 d) 11 e) 924924 f) 88

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