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A) Establish that nCr = n! / r!(n-r)! (b) Show that 8C5 = 8C3 (c) Find the values of (i) 14!/(10!x4!) (ii) 9!/(6!x3!) (iii) 20!/(10!x10!)

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Answers

3 parts
c(i)

1001\text{1001}

c(ii)

84\text{84}

c(iii)

184756\text{184756}

Step-by-step solution

15 steps

Part a· 3 steps

To establish the formula for nCrnC_r

  1. Define combinations.

    A combination nCrnC_r represents the number of ways to choose rr items from a set of nn distinct items, where the order of selection does not matter.

  2. Relate to permutations.

    The number of permutations of rr items chosen from nn is given by nPr=n!(n−r)!nP_r = \frac{n!}{(n-r)!}. Since the order of the rr chosen items does not matter in a combination, we divide the number of permutations by the number of ways to arrange the rr items, which is r!r!.

  3. State the formula.

    Therefore, the formula for nCrnC_r is:

    nCr=nPrr!=n!r!(n−r)!nC_r = \frac{nP_r}{r!} = \frac{n!}{r!(n-r)!}

Part b· 3 steps

To show that 8C5=8C38C_5 = 8C_3

  1. Calculate 8C58C_5 using the formula nCr=n!r!(n−r)!nC_r = \frac{n!}{r!(n-r)!}.

    8C5=8!5!(8−5)!=8!5!3!8C_5 = \frac{8!}{5!(8-5)!} = \frac{8!}{5!3!}

  2. Calculate 8C38C_3 using the formula.

    8C3=8!3!(8−3)!=8!3!5!8C_3 = \frac{8!}{3!(8-3)!} = \frac{8!}{3!5!}

  3. Compare the results.

    Since 5!3!=3!5!5!3! = 3!5!, we can see that:

    8!5!3!=8!3!5!\frac{8!}{5!3!} = \frac{8!}{3!5!}

    Thus, 8C5=8C38C_5 = 8C_3.

Part c(i)

1001\text{1001}

14!10!×4!\frac{14!}{10! \times 4!}

To find the values of the expressions

  1. Expand the factorials and simplify.

    14!10!×4!=14×13×12×11×10!10!×(4×3×2×1)\frac{14!}{10! \times 4!} = \frac{14 \times 13 \times 12 \times 11 \times 10!}{10! \times (4 \times 3 \times 2 \times 1)}

  2. Cancel $10!$ and calculate the remaining terms.

    =14×13×12×114×3×2×1=14×13×12×1124= \frac{14 \times 13 \times 12 \times 11}{4 \times 3 \times 2 \times 1} = \frac{14 \times 13 \times 12 \times 11}{24}

  3. Perform the multiplication and division.

    =2402424=1001= \frac{24024}{24} = 1001

    The value is 1001\boxed{\text{1001}}.

Part c(ii)

84\text{84}

9!6!×3!\frac{9!}{6! \times 3!}

  1. Expand the factorials and simplify.

    9!6!×3!=9×8×7×6!6!×(3×2×1)\frac{9!}{6! \times 3!} = \frac{9 \times 8 \times 7 \times 6!}{6! \times (3 \times 2 \times 1)}

  2. Cancel $6!$ and calculate the remaining terms.

    =9×8×73×2×1=9×8×76= \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = \frac{9 \times 8 \times 7}{6}

  3. Perform the multiplication and division.

    =5046=84= \frac{504}{6} = 84

    The value is 84\boxed{\text{84}}.

Part c(iii)

184756\text{184756}

20!10!×10!\frac{20!}{10! \times 10!}

  1. Expand the factorials and simplify.

    20!10!×10!=20×19×18×17×16×15×14×13×12×11×10!10!×(10×9×8×7×6×5×4×3×2×1)\frac{20!}{10! \times 10!} = \frac{20 \times 19 \times 18 \times 17 \times 16 \times 15 \times 14 \times 13 \times 12 \times 11 \times 10!}{10! \times (10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)}

  2. Cancel $10!$ and simplify the remaining terms by cancelling common factors.

    =20×19×18×17×16×15×14×13×12×1110×9×8×7×6×5×4×3×2×1= \frac{20 \times 19 \times 18 \times 17 \times 16 \times 15 \times 14 \times 13 \times 12 \times 11}{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} =(20/(10×2))×(18/9)×(16/8)×(15/(5×3))×(14/7)×(12/6)×19×17×13×11×(1/4)= (20/(10 \times 2)) \times (18/9) \times (16/8) \times (15/(5 \times 3)) \times (14/7) \times (12/6) \times 19 \times 17 \times 13 \times 11 \times (1/4) =1×2×2×1×2×2×19×17×13×11×(1/4)= 1 \times 2 \times 2 \times 1 \times 2 \times 2 \times 19 \times 17 \times 13 \times 11 \times (1/4) =16×19×17×13×114= \frac{16 \times 19 \times 17 \times 13 \times 11}{4}

  3. Perform the multiplication and division.

    =4×19×17×13×11= 4 \times 19 \times 17 \times 13 \times 11 =4×323×143= 4 \times 323 \times 143 =4×46189= 4 \times 46189 =184756= 184756

    The value is 184756\boxed{\text{184756}}.

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