Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

Arrangement question?

For question 25

I have no idea to deal with, I used 1x3x2x1x1 to solve but the answer is only 130

For question 26

Actually I don’t really understand the question. Why does it mean which is painted in one??? I just can find the answer for a which is 360. For b I tried to use (6C2x6C2)/(2!x2!) but the answer is only 6. The answers for c and d are 12 and 1170

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  • 1 month ago
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    25.

    1-digit even numbers: 4 and 6

    No. of 1-digit even numbers = 2

    2-digit even numbers:

    The units digit is "4" or "6" (₂P₁). Assign 1 of the rest 4 digits to the tens digit place (₄P₁).

    No. of 2-digit even numbers = ₂P₁ × ₄P₁ = 2 × 4 = 8

    3-digit even numbers:

    The units digit is "4" or "6" (₂P₁). Assign 2 of the rest 4 digits to the rest 2 digit places (₄P₂).

    No. of 3-digit even numbers = ₂P₁ × ₄P₂ = 2 × 12 = 24

    4-digit even numbers:

    The units digit is "4" or "6" (₂P₁). Assign 3 of the rest 4 digits to the rest 3 digit places (₄P₃).

    No. of 3-digit even numbers = ₂P₁ × ₄P₃ = 2 × 24 = 48

    5-digit even numbers:

    The units digit is "4" or "6" (₂P₁). Assign the rest 4 digits to the rest 4 digit places (₄P₄).

    No. of 3-digit even numbers = ₂P₁ × ₄P₄ = 2 × 24 = 48

    Total no. of even numbers formed

    = 2 + 8 + 24 + 48 + 48

    = 130

    ====

    26.

    (a)

    There are 1 red peg, 1 white peg, 1 black peg, 1 green peg, 1 blue peg and 1 yellow peg.

    Out of the 6 pegs, place 4 pegs in the 4 holes (₆P₄).

    Number of arrangements = ₆P₄ = 360

    ----

    (b)

    Placing 4 pegs in the 4 holes is ₄P₄. Divided by 2!2! because there are 2 red pegs and 2 white pegs.

    No. of arrangements = ₄P₄/(2!2!) = 24/4 = 6

    ----

    (c)

    Placing 4 pegs into the 4 holes is ₄P₄. Divided by 2! because there are 2 red pegs.

    No. of arrangements = ₄P₄/(2!2!) = 24/4 = 6

    ----

    (d)

    There are 2 red pegss, 2 white pegs, 2 black pegs, 2 green pegs, 2 blue pegs and 2 yellow pegs.

    4 pegs with 2 colours:

    Firstly, choose the 2 colours from the 6 colours(₆C₂). Placing 4 pegs into the 4 holes is ₄P₄. Divided by 2!2! because each pair of pegs are painted with the same colour.

    No of arrangements for 4 pegs with 2 colours

    = ₆C₂ × ₄P₄ / (2!2!)

    = 15 × 24 / 4

    = 90

    4 pegs with 3 colours:

    Firstly, choose 1 colour from the 6 colours for 2 pegs (₆C₁). Then, choose other 2 colours from the rest 5 colours for other 2 pegs (₅C₂). Placing 4 pegs into the 4 holes is ₄P₄. Divided by 2! because one pair of pegs are painted with the same colour.

    No of arrangements for 4 pegs with 3 colours

    = ₆C₁ × ₅C₂ × ₄P₄ / 2

    = 6 × 10 × 24 / 2

    = 720

    4 pegs with 4 colours:

    Choose 4 colours from the 6 colours for 4 pegs (₆C₄). Placing 4 pegs into the 4 holes is ₄P₄

    No of arrangements for 4 pegs with 4 colours

    = ₆C₄ × ₄P₄

    = 15 × 24

    = 360

    Total no. of arrangements

    = 90 + 720 + 360

    = 1170

  • User
    Lv 7
    1 month ago

    4 and 6 are the only even numbers

    and

    the final digit of an even number is always an even number.

    SO: you have

    single digit - 2 even numbers

    double digit - 4 x 2 even numbers

    etc.

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