Flashcards · 12 cards

Permutation & combination

Say the answer to yourself, then check. Cards you know come back less and less often; cards you don't come back tomorrow.

  1. Know it

    nPr{}^nP_r and nCr{}^nC_r?

    Answer

    nPr=n!(n−r)!{}^nP_r = \dfrac{n!}{(n - r)!} (order matters) and nCr=n!r! (n−r)!{}^nC_r = \dfrac{n!}{r!\,(n - r)!} (order doesn't).

  2. Rule

    How are nPr{}^nP_r and nCr{}^nC_r related?

    Answer

    nPr=nCr×r!{}^nP_r = {}^nC_r \times r!: each selection of rr things gives r!r! arrangements.

    selectionsarrangementsA, BABBAA, CACCAB, CBCCB³C₂ = 3 selections, × 2! = 6 = ³P₂ arrangements
    Arrangements and selectionsEach selection of r gives r! arrangements: ⁿPᵣ = ⁿCᵣ × r!
  3. Know it

    Solve nC2=45{}^nC_2 = 45.

    Answer

    Write it out: n(n−1)2=45\dfrac{n(n - 1)}{2} = 45, so n2−n−90=0n^2 - n - 90 = 0 and n=10n = 10 (not −9-9).

  4. Rule

    Some items must stand together. How do you count the arrangements?

    Answer

    Glue them into one block. Arrange the blocks and single items, then arrange inside each block, and multiply.

    ABCDE4 units: 4! = 24 ordersCD or DC inside: × 2! = 48
    Glue them togetherArrange the units, then arrange inside each block
  5. Rule

    Arranging letters when some repeat?

    Answer

    n!p! q!⋯\dfrac{n!}{p!\,q!\cdots}: divide by the factorial of each repeat count (three A's: divide by 3!=63! = 6, not 3).

    A's tagged: 3! = 6tags offA₁A₂BA₂A₁BAABA₁BA₂A₂BA₁ABABA₁A₂BA₂A₁BAAeach word counted 2! times: 3! ÷ 2! = 3 words
    Take the tags offA, A, B: 3! orders with the A's tagged, but each word appears 2! times, so 3!/2! = 3
  6. Know it

    One place is restricted, such as the last digit being even. Where do you start?

    Answer

    Fill the restricted place first, then the others.

  7. Rule

    A committee must have at least 2 women. How do you count?

    Answer

    Count each case (2 women, 3 women, …) separately, then add.

    3 people, at least 2 women (4 women, 5 men)2 W + 1 M⁴C₂ × ⁵C₁ = 30+3 W⁴C₃ = 430 + 4 = 34 committeeslist the cases that fit, count each, then add
    At least 2 womenCount each case, then add
  8. Rule

    Splitting nn people into kk equal, unnamed groups of mm?

    Answer

    n!(m!)k k!\dfrac{n!}{(m!)^k\,k!}: divide by k!k! because swapping whole groups gives the same split.

    pair 1, pair 2same splitABCD=CDABACBD=BDACADBC=BCAD4! ÷ (2! 2!) = 6 numbered, ÷ 2! = 3 splits
    Unnamed pairsA, B, C, D into two pairs: numbering the pairs counts each split 2! times
  9. Know it

    Counting: when do you multiply and when do you add?

    Answer

    Choosing from one group and another: multiply. One case or another: add.

  10. Know it

    A probability from selections?

    Answer

    number of selections that fittotal number of selections\dfrac{\text{number of selections that fit}}{\text{total number of selections}}.

  11. Which method?

    WAEC 2016 · Paper 2 · Q13 (a)

    Two different Mathematics books, 5 different Physics books and 3 different Chemistry books are to be arranged on a shelf. How many arrangements are possible if: (i) books on the same subject must stand together; (ii) only the Physics books must stand together?

    How do you keep each subject together?

    Answer

    Glue each subject into a block: 3!3! ways to order the three blocks, times 2!×5!×3!2! \times 5! \times 3! for the orders inside them.

  12. Which method?

    WAEC 2011 · Paper 2 · Q13 (a)

    A committee of five is to be formed among 6 Ghanaians, 8 Nigerians and 5 Gambians. In how many ways can the committee be formed if: (i) there is no restriction; (ii) at most 2 Ghanaians are on the committee; (iii) 1 Nigerian is on the committee?

    "At most 2 Ghanaians": which cases?

    Answer

    0, 1 or 2 Ghanaians, with the rest chosen from the 13 others. Count each case and add.