Tiêu đề Permutations & Combinations CQ & MCQ Practice Sheet Solution (HSC 26).pdf ✅

2  Higher Math 1st Paper Chapter-5 3| c`v_©weÁvbx Issac Newton 1643 mv‡j England G Rb¥MÖnY K‡ib| [Kz. †ev. 17] (K) 1, 2, 3, 4, 5, 6, 7, 8, 9 AsK ̧‡jv n‡Z 3 Øviv wefvR ̈ bq Giƒc AsK ̧‡jvi ga ̈ n‡Z 4 wU K‡i AsK KZfv‡e evQvB Kiv hvq? (L) †`LvI †h, `„k ̈K‡íi weÁvbxi bv‡gi †kl As‡ki web ̈vm msL ̈v Zvi bv‡gi cÖ_g As‡ki web ̈vm msL ̈v 12 ̧Y| (M) `„k ̈K‡í weÁvbxi Rb¥f~wgi bv‡gi eY© ̧wj †Z‡K cvuPwU K‡i eM© wb‡q KZ ̧wj web ̈vm MVb Kiv hvq? mgvavb: (K) 3 Øviv wefvR ̈ bq Giƒc AsK ̧‡jv 1, 2, 4, 5, 7, 8 A_©vr †gvU 6 wU  6 wU wfbœ AsK n‡Z 4 wU evQvB Kivi Dcvq msL ̈v 6C4 ev 15 (L) †kl Ask Newton G †gvU eY© 6 wU  web ̈vm msL ̈v = 6! = 720 1g Ask Issac G †gvU eY© 5 wU hv‡`i g‡a ̈ s Av‡Q 2wU  web ̈vm msL ̈v = 5! 2! = 60  †kl As‡ki web ̈vm = 720 = 12  60 = 12  1g As‡ki web ̈vm  `„k ̈K‡íi weÁvbxi bv‡gi †kl As‡ki web ̈vm msL ̈v Zvi bv‡gi cÖ_g As‡ki web ̈vm msL ̈vq 12 ̧Y| (M) England kãwU‡Z †gvU 7 wU eY© Av‡Q| hvi g‡a ̈ 2 wU n Ges evKx me wfbœ wfbœ| eY© ̧wj n‡”Q, E, n, n, g, l, a, d. ca`Ë kãwU n‡Z 5 wU K‡i eY© wb‡q web ̈vm msL ̈v: (i) 2wU n I wZbwU wfbœ eY© wb‡q = 5C3  5! 2! (ii) me ̧‡jv wfbœ wfbœ eY© wb‡q = 6C5  5!  †gvU MwVZ k‡ãi msL ̈v =    5C  3  5! 2! + 6C5  5! = 600 + 720 = 1320 4| GKwU GKw`‡bi g ̈v‡P NEWZEALAND wμ‡KU `‡j 7 Rb e ̈vUmg ̈vb, 6 Rb †evjvi Ges 2 Rb DB‡KU wKcvi ivLv n‡jv| [w`. †ev. 17] (K) nP3 = 2  nC4 n‡j n Gi wbY©q Ki| (L) DÏxcK n‡Z KZfv‡e 11 Rb †L‡jvqv‡oi `j MVb Kiv hv‡e hv‡Z me©`v 5 Rb †evjvi Ges Kgc‡ÿ GKRb DB‡KU wKcvi _vK‡e? (M) ̄^ieY© ̧‡jv‡K cvkvcvwk bv †i‡L NEWZEALAND k‡ãi Aÿi ̧‡jv‡K KZ cÖKv‡i mvRv‡bv hv‡e? mgvavb: (K) †`Iqv Av‡Q, n P3 = 2  nC4  n! (n – 3)! = 2  n! 4!(n – 4)!  n! (n – 3) (n – 4)! = 2  n! 4! (x – 4)!  1 n – 3 = 1 12  n – 3 = 12  n = 15 (L) `j MVb cÖwμqv wb¤œiƒc: e ̈vUmg ̈vb (7) †evjvi (6) DB‡KU wKcvi (2) (i) 5 5 1 (ii) 4 5 2  `j MV‡bi †gvU Dcvq = 7C5  6C5  2C1 + 7C4  6C5  2C2 = 462 (M) NEWZEALAND k‡ãi †gvU Aÿi 10 wU Aÿi Av‡Q hvi g‡a ̈ 2 wU E, 2 wU A, 2 wU N we` ̈gvb Ges evwK eY© ̧wj wfbœ wfbœ|  †gvU mvRv‡bv msL ̈v = 10! 2! 2! 2! = 453600 ̄^ieY© 4 wU (E, E, A, A) †K 1 wU Aÿi g‡b K‡i †gvU mvRv‡bv msL ̈v = 7! 2! = 2520 Avevi, ̄^ieY© 4 wU wb‡R‡`i g‡a ̈ mvRv‡bv msL ̈v = 4! 2! 2! = 6  ̄^ieY© ̧‡jv‡K cvkvcvwk †i‡L mvRv‡bv msL ̈v = 2520  6 = 15120  ̄^ieY© ̧‡jv‡K cvkvcvwk bv †i‡L mvRv‡bv msL ̈v, = 453600 – 15120 = 438480 5| `„k ̈Kí-1: mv`vZ 0, 3, 4, 5, 6, 9 AsK ̧wj wjL‡Z cv‡i| `„k ̈Kí-2:  A = 3  i + 2  j + 6  k Ges  B =  i – 4  j – 3  k [mKj †ev. 18] (K) hw` nP2 = 3 nC3 nq Z‡e n Gi gvb wbY©q Ki| (L)  A eivei  B Gi Dcvsk wbY©q Ki| (M) `„k ̈Kí-1 n‡Z, cÖ‡Z ̈K A1⁄4‡K cÖ‡Z ̈K msL ̈vq †Kej GKevi e ̈envi K‡i Qq A1⁄4wewkó KZ ̧wj A_©c~Y© †Rvo msL ̈v MVb Kiv hvq Zv wbY©q Ki| mgvavb: (K) n P2 = 3  nC3  n! (n – 2)! = 3  n! 3!(n – 3)!  1 (n – 2)! = 3  1 6(n – 3)!  1 (n – 2) (n – 3)! = 1 2(n – 3)!  1 n – 2 = 1 2  n – 2 = 2  n = 4 (L)  A = 3  i + 2  j + 6  k ,  B =  i – 4  j – 3  k  A eivei  B Gi Dcvsk = Bcos  a ev  A .  B A  a
web ̈vm I mgv‡ek  CQ & MCQ Practice Sheet Solution (HSC 26) 3 GLb, A = 9 + 4 + 36 = 49 = 7 B = 1 + 16 + 9 = 26  A .  B = 3  1 + 2  (– 4) + 6(– 3) = 3 – 8 – 18 = – 23  wb‡Y©q Dcvsk =  A .  B A  a = – 23 7  A A = – 23 7 3  i + 2  j + 6  k 7 = – 23 49 (3 )  i + 2  j + 6  k (M) cÖ`Ë A1⁄4 ̧wj: 0, 3, 4, 5, 6, 9 †Rvo msL ̈vi Rb ̈ GK‡Ki ̄’v‡b 6 A_ev 4 A_ev 0 _vK‡e|  GK‡Ki ̄’v‡b 6 A_ev 4 A_ev 0 Ges Aewkó 5wU A1⁄4 wb‡q MwVZ msL ̈vi cwigvY = 3C1  5! = 360 Avevi, GK‡Ki ̄’v‡b 6 A_ev 4 Ges ïiæ‡Z 0 wb‡q MwVZ Qq A‡1⁄4i msL ̈vi cwigvY = 2C1  4! = 48  A1⁄2 ̧‡jv GKevi e ̈envi K‡i A_©c~Y© †Rvo msL ̈v = 360 – 48 = 312 6| BANGLADESH Avgv‡`i gvZ...f~wg| (K) nP3 = 210 n‡j n Gi gvb KZ? (L) ̄^ieY© ̧wj GK‡Î bv †i‡L DÏxc‡Ki Bs‡iwR kãwUi Aÿi ̧wj‡K KZ Dc‡q mvRv‡bv hvq? (M) Bs‡iwR kãwU n‡Z 4 wU K‡i eY© wb‡q KZ ̧wj kã MVb Kiv hvq? mgvavb: (K) n P3 = 210  n(n – 1) (n – 2) = 210  (n2 – n) (n – 2) = 210  n 3 – 2n2 + 2n = 210  n 3 – 3n2 + 2n – 210 = 0  n 3 – 7n2 + 4n2 – 28n + 30n – 210 = 0  n 2 (n – 7) + 4n(n – 7) + 30(n – 7) = 0  (n – 7) (n2 + 4n + 30) = 0  n – 7 = 0 A_ev n 2 + 4n + 30 = 0  n = 7 A_ev n = – 4  16 – 120 2 (MÖnY‡hvM ̈ bq)  n = 7 Gi Rb ̈ n P3 = 210 n‡e| (L) BANGLADESH kãwU‡Z †gvU 10 wU Aÿi Av‡Q| G‡`i g‡a ̈ 2 wU A Av‡Q Ges evwK ̧‡jv wfbœ wfbœ|  kãwUi Aÿi ̧wji web ̈vm msL ̈v = 10! 2! = 1814400 Avevi, kãwU‡Z 3 wU ̄^ieY© I 7 wU e ̈ÄbeY© Av‡Q| ̄^ieY© 3 wU †K GKwU Aÿi g‡b K‡i †gvU 8 wU| Aÿi wb‡q web ̈vm msL ̈v = 8! = 40320 Avevi, ̄^ieY© 3 wU‡K wb‡R‡`i g‡a ̈ mvRv‡bv hvq 3! 2! ev 3 Dcv‡q|  ̄^ieY© ̧wj‡K GK‡Î †i‡L web ̈vm msL ̈v = 40320  3 = 120960  ̄^ieY© ̧wj‡K GK‡Î bv †i‡L web ̈vm msL ̈v = 1814400 – 120960 = 1693440 (M) BANGLADESH kãwU‡Z †gvU Aÿi Av‡Q 10 wU| G‡`i g‡a ̈ 2 wU A Av‡Q| evwK ̧‡jv wfbœ wfbœ| A_©vr AA B N G L D E S H kãwU n‡Z 4 wU e‡Y©i evQvB I mvRv‡bv msL ̈v wb¤œiƒc: eY©bv mgv‡ek web ̈vm (i) `ywU GKB Aÿi, `ywU wfbœ Aÿi 1  8C2 1  8C2  4! 2! (ii) 4 wU wfbœ Aÿi 9C4 9C4  4!  wb‡Y©q kã MVb msL ̈v = 1  8C2  4! 2! + 9C4  4! = 336 + 3024 = 3360 7| „Practice sharpens knowledge‟ [Kz. †ev. 19] (K) GKwU cÂfz‡Ri K‡Y©i msL ̈v wbY©q Ki| (L) DÏxc‡Ki †kl kãwUi eY© ̧wji mvRv‡bv e ̈e ̄’vq KZ ̧wj‡Z ̄^ieY© ̧wj GK‡Î _vK‡e bv? (M) DÏxc‡Ki cÖ_g kãwUi eY© ̧wj †_‡K cÖwZev‡i 4 wU eY© w`‡q KZ ̧wj kã MVb Kiv hvq? mgvavb: (K) GKwU cÂfz‡Ri cvuPwU †KŠwYK we›`y we` ̈gvb| GB cvuP †KŠwYK we›`yi †h‡Kv‡bv `yB †KŠwYK we›`yi ms‡hv‡M MwVZ †iLvi msL ̈v = 5C2 hvi g‡a ̈ 5 wU cvk¦© †iLv KY© bq|  wb‡Y©q K‡Y©i msL ̈v = 5C2 – 5 = 5 (L) knowledge kãwU‡Z †gvU 9 wU eY© Av‡Q| hvi g‡a ̈ wZbwU ̄^ieY© (o, e, e) Ges QqwU e ̈ÄbeY© (k, n, w, l, d, g)  9 wU eY© mvRv‡bv msL ̈v = 9! 2! GLb, ̄^ieY© ̧‡jv‡K GK‡Î GKwU Aÿi a‡i mvRv‡bvi msL ̈v = 7! Avevi ̄^ieY© ̧‡jv‡K wb‡R‡`i g‡a ̈ mvRv‡bv hvq 3! 2!  ̄^ieY© ̧‡jv‡K GK‡Î †i‡L mvRv‡bv msL ̈v = 7!  3! 2!  ̄^ieY© ̧‡jv GK‡Î bv †i‡L mvRv‡bv msL ̈v = 9! 2! – 7!  3! 2! = 166320 (M) Practice kãwU‡Z †gvU 8 wU eY© Av‡Q hvi g‡a ̈ 2 wU c Ges evwK ̧‡jv wfbœ wfbœ| cÖwZev‡i PviwU K‡i eY© wb‡q kã MV‡bi Dcvq: (i) 2 wU c I evwK `ywU wfbœ eY© wb‡q web ̈vm msL ̈v = 6C2  4! 2!