Before you begin

Twelve ques­tions to fin­ish the chap­ter. Before you write a sin­gle for­mula, ask your­self one ques­tion: does order mat­ter here? If it does, you want a per­mu­ta­tion; if not, a com­bi­na­tion. After that, deal with any restric­tion first (a fixed let­ter, peo­ple who must sit together) and only then count the rest.

Ques­tions

  1. How many 44-digit even num­bers can be formed from 1,2,3,4,5,6,71, 2, 3, 4, 5, 6, 7 with­out rep­e­ti­tion?
  2. In how many ways can 77 peo­ple sit in a row if two par­tic­u­lar peo­ple refuse to sit together?
  3. How many arrange­ments of ENGI­NEER­ING are there? How many start and end with E?
  4. Find rr if 5×4Pr=6×5Pr−15 \times {}^4P_r = 6 \times {}^5P_{r-1}.
  5. How many 33-let­ter words (with or with­out mean­ing) can be made from COM­PUTER with­out rep­e­ti­tion? How many con­tain the let­ter M?
  6. How many ways can 33 prizes be given to 1010 stu­dents if a stu­dent may win more than one prize? If not?
  7. A cricket team of 1111 is cho­sen from 66 bats­men, 55 bowlers and 33 all-rounders, with exactly 55 bats­men, 44 bowlers and 22 all-rounders. How many ways?
  8. 1212 points lie in a plane; 55 of them are collinear and no other three are. How many lines? How many tri­an­gles?
  9. How many 55-card hands have exactly 33 hearts?
  10. A ques­tion paper has two sec­tions of 66 ques­tions each. A stu­dent must answer 77 ques­tions, with at least 33 from each sec­tion. In how many ways can she choose?
  11. How many words can be formed from all let­ters of MATH­E­MAT­ICS with the vow­els together?
  12. Show that nPr=r⋅nCr⋅(r−1)!^nP_r = r \cdot {}^{n}C_r \cdot (r - 1)! and hence that nPr≥nCr^nP_r \ge {}^nC_r.
Twelve points in a plane for question 8: five orange points lie on one straight line, and seven blue points are scattered above it with no three of all the points in a line otherwise.
Ques­tion 8: twelve points, five of them on one line

Answers

Show answers
  1. Last digit 2,42, 4 or 66: 3×6P3=3×120=3603 \times {}^6P_3 = 3 \times 120 = 360.
  2. Count all seat­ings and sub­tract those where the two sit together: 7!−2×6!=5040−1440=36007! - 2 \times 6! = 5040 - 1440 = 3600.
  3. E3, N3, G2, I2, R1: 11!3! 3! 2! 2!=277,200\displaystyle \tfrac{11!}{3!\,3!\,2!\,2!} = 277{,}200; with E at both ends: 9!3! 2! 2!=15,120\displaystyle \tfrac{9!}{3!\,2!\,2!} = 15{,}120.
  4. 5⋅4!(4−r)!=6⋅5!(6−r)!\displaystyle 5 \cdot \tfrac{4!}{(4 - r)!} = 6 \cdot \tfrac{5!}{(6 - r)!} gives (6−r)(5−r)=6(6 - r)(5 - r) = 6, r=3r = 3.
  5. 8P3=336^8P_3 = 336; con­tain­ing M: 3×7P2=1263 \times {}^7P_2 = 126.
  6. 10001000; 720720.
  7. 6C5 5C4 3C2=6×5×3=90^6C_5\,{}^5C_4\,{}^3C_2 = 6 \times 5 \times 3 = 90.
  8. Lines 12C2−5C2+1=57^{12}C_2 - {}^5C_2 + 1 = 57; tri­an­gles 12C3−5C3=210^{12}C_3 - {}^5C_3 = 210.
  9. 13C3 39C2=286×741=211,926^{13}C_3\,{}^{39}C_2 = 286 \times 741 = 211{,}926.
  10. (3,4)(3, 4) or (4,3)(4, 3): 2×6C3 6C4=2×20×15=6002 \times {}^6C_3\,{}^6C_4 = 2 \times 20 \times 15 = 600.
  11. Vow­els A, E, A, I as one block (4!2!=12\displaystyle \tfrac{4!}{2!} = 12 orders); with M2, T2, H, C, S: 88 units, 8!2! 2!=10,080\displaystyle \tfrac{8!}{2!\,2!} = 10{,}080; total 120,960120{,}960.
  12. nPr=r! nCr=r⋅(r−1)! nCr^nP_r = r!\,{}^nC_r = r \cdot (r - 1)!\,{}^nC_r; and r!≥1r! \ge 1.