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

web ̈vm I mgv‡ek  CQ & MCQ Practice Sheet Solution (HSC 26) 1 05 web ̈vm I mgv‡ek Permutations & Combinations WRITTEN 1| m¤úawZ wWwR Awdm n‡Z PDS dvBj Avc-Uz †WU Kivi Rb ̈ cÖwZwU K‡j‡R wb‡`©k †`q| wb‡`©kgZ Avgv‡`i mnKg©x wgt Lvb Zvi BDRvi AvBwW “COMBINATION” Ges cvmIqvW© “10652” e ̈envi K‡i| [Xv. †ev. 17] (K) nPr = 54 Ges nCr = 9 n‡j, r Gi gvb wbY©q Ki| (L) BDRvi AvBwW Gi eY© ̧wj n‡Z cÖwZevi PviwU K‡i eY© wb‡q KZ Dcv‡q mvRv‡bv hv‡e? (M) cvmIqv‡W©i cÖ‡Z ̈K AsK‡K cÖwZ msL ̈vq †Kej GKevi e ̈envi K‡i cvuP AsK wewkó KZ ̧wj A_©c~Y© †e‡Rvo msL ̈v MVb Kiv hvq? mgvavb: (K) n Pr = 54  n! (n – r)! = 54 ...... (i) nCr = 9  n! r!(n – r)! = 9 ...... (ii) (i)  (ii)  r! = 6  r! = 3!  r = 3 (L) „COMBINATION‟ kãwU‡Z 11 wU eY© Av‡Q †h ̧wj wb¤œiƒc: OO NN II C M B A T 4 wU eY© wb‡q mvRv‡bv Dcvq: (i) 3 †Rvov GKB eY© n‡Z 2 †Rvov wb‡q = 3C2  4! 2!  2! (ii) 3 †Rvov GKB eY© n‡Z 1 †Rvov I evwK `yBwU wfbœ eY© wb‡q = 3C1  7C2  4! 2! (iii) 4 wU wfbœ wfbœ eY© w`‡q = 8C4  4!  wb‡Y©q mvRv‡bv msL ̈v = 3C2  4! 2!  2! + 3C1  7C2  4! 2! + 8C4  4! = 18 + 756 + 1680 = 2454 (M) cÖ`Ë A1⁄4 ̧wj 1, 0, 6, 5, 2 we‡Rvo msL ̈vi Rb ̈ GK‡Ki ̄’v‡b 5 A_ev 1 n‡e|  GK‡Ki ̄’v‡b 5 A_ev 1 wb‡q Aewkó 4 wU A1⁄4 Øviv MwVZ msL ̈vi cwigvY = 2C1  4! = 48 Avevi ïiæ‡Z 0 Ges †k‡l 1 ev 5 wb‡q MwVZ A1⁄4 ̧‡jvi Øviv MwVZ msL ̈vi cwigvY = 2C1  3! = 12  A1⁄4 ̧‡jv Øviv cvuP A‡1⁄4i A_©c~Y© we‡Rvo msL ̈v = 48 – 18 = 36 wU 2| GKwU K¬v‡ei Kvh©wbe©vnx KwgwUi m`m ̈ 21 Rb, hvi g‡a ̈ 8 Rb gwnjv I 13 Rb cyiæl| [P. †ev. 17] (K) DEPRESSION kãwUi Aÿi ̧wj‡K KZfv‡e mvRv‡bv hv‡e hv‡Z ̄^ieY© ̧wj GK‡Î _vK‡e? (L) GKRb gwnjv m`m ̈ mfvcwZ n‡j mfvcwZ‡K ev` w`‡q 11 m`m ̈wewkó KZ ̧‡jv DcKwgwU MVb Kiv hv‡e hv‡Z Kgc‡ÿ 4 Rb gwnjv m`m ̈ AšÍfz©3 _vK‡e? (M) Kvh©wbe©vnx KwgwU n‡Z `yB Rb cyiæl ev` w`‡q Aewkó m`m ̈e„›`‡K GK mvwi‡Z KZfv‡e mvRv‡bv hv‡e hv‡Z `yBRb gwnjv m`m ̈ GK‡Î bv _v‡K| mgvavb: (K) DEPRESSION kãwU‡Z †gvU eY© 10 wU hv‡Z ̄^ieY© Av‡Q 4 wU (E, E, I, O)| ̄^ieY© 4 wU 1 wU eY© a‡i †gvU eY© 7 wU hv‡Z S Av‡Q 2 wU|  7 wU eY©‡K wb‡R‡`i g‡a ̈ web ̈vm msL ̈v 7! 2!  ̄^ieY© ̧wj GK‡Î _v‡K Giƒc web ̈vm msL ̈v = 7! 2!  4! 2! (L) 11 Rb KwgwU MV‡bi Rb ̈ m`m ̈ wbe©vPb I KwgwU MVb wb¤œiƒc n‡e| KwgwU MV‡bi Dcvq cyiæl (13 Rb) gwnjv (mfvcwZ ev‡` 7 Rb) (i) 7 4 13C7  7C4 (ii) 6 5 13C6  7C5 (iii) 5 6 13C5  7C6 (iv) 4 7 13C4  7C7 KwgwU MV‡bi †gvU Dcvq = (13C7  7C4 + 13C6  7C5 + 13C5  7C6 + 13C4  7C7) = 60,060 + 36,036 + 9,009 + 715 = 1,05,820 (M) 2 Rb cyiæl ev‡` Aewkó cyiæl (13 – 2) Rb ev 11 Rb| 11 Rb cyiæ‡li gv‡S 8 Rb gwnjv‡K mvRv‡Z n‡e †hb 2 Rb gwnjv m`m ̈ GK‡Î bv _v‡K|  11 Rb cyiæ‡li gv‡S 10 wU Ges `yB cÖv‡šÍ 2 wU mn †gvU 12 wU ̄’v‡b 8 wU gwnjv‡K ivLvi web ̈vm msL ̈v = 12P8 Avevi, 11 Rb cyiæl‡K wb‡R‡`i g‡a ̈ web ̈vm msL ̈v = 11!  wb‡Y©q mvRv‡bvi msL ̈vi = 12P8  11!

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!
4  Higher Math 1st Paper Chapter-5 (ii) 4 wU wfbœ eY© wb‡q web ̈vm msL ̈v = 7C4  4!  †gvU 4 wU e‡Y©i k‡ãi msL ̈v = 6C2  4! 2! + 7C4  4! = 180 + 840 = 1020 8| Examination kãwU‡Z 11 wU eY© Av‡Q| (K) nC8 = nC3 n‡j nP6 Gi gvb KZ? (L) DÏxc‡Ki Bs‡iwR kãwUi e ̈ÄbeY© ̧wj‡K GK‡Î bv †i‡L KZ iK‡g mvRv‡bv hvq? (M) DÏxc‡Ki Bs‡iwR kãwUi eY© ̧wj †_‡K 5 wU K‡i eY© wb‡q mgv‡ek msL ̈v wbY©q Ki| mgvavb: (K) nC8 = nC3 n‡j, n = 8 + 3 = 11  n P6 = 11P6 = 332640 (L) Examination kãwU †gvU 11 wU Aÿi Av‡Q| hvi g‡a ̈ 6 wU ̄^ieY© (E, a, a, i, i, o) Ges 5 wU e ̈ÄbeY© (x, m, n, n, t)  kãwUi me ̧‡jv Aÿi GK‡Î wb‡q web ̈vm msL ̈v = 11! 2! 2! 2! = 4989600 e ̈ÄbeY© 5 wU †K GKwU Aÿi a‡i web ̈vm msL ̈v = 7! 2! 2! = 1260 Avevi, e ̈ÄbeY© ̧wj‡K wb‡R‡`i g‡a ̈ mvRv‡bv msL ̈v = 5! 2! = 60  e ̈ÄbeY© ̧wj‡K GK‡Î †i‡L web ̈vm msL ̈v = 1260  60 = 75600  e ̈ÄbeY© ̧wj‡K GK‡Î †i‡L web ̈vm msL ̈v = 4989600 – 75600 = 4914000 (M) Examination kãwU‡Z †gvU 11 wU eY© Av‡Q| hvi g‡a ̈ 2 wU a, 2 wU i Ges 2 wU n Av‡Q| evwK ̧wj wfbœ wfbœ| kãwUi Aÿi ̧wj n‡Z 5 wU eY© wb‡q evQvB‡qi Dcvq: (i) `yB †Rvov GKB eY© Ges evwKwU wfbœ wb‡q = 3C2  6C1 (ii) `yBwU GKB Ges evwK wZbwU wfbœ wfbœ wb‡q = 3C1  7C3 (iii) ca‡Z ̈KwU eY© wfbœ wb‡q = 8C5  wb‡Y©q mgv‡ek msL ̈v = 3C2  6C1 + 3C1  7C3 + 8C5 = 18 + 105 + 56 = 179 9| „PERMUTATION‟ kãwUi A_© web ̈vm| (K) MÖvgxY‡dv‡bi †gvevBj b¤^i ̧wj 11 A‡1⁄4i hvi g‡a ̈ cÖ_g wZbwU †KvW b¤^i 017 n‡j †gvU KZ ̧wj ms‡hvM Pvjy Ki‡Z cvi‡e? (L) ̄^ieY© ̧wj‡K †Rvo ̄’v‡b †i‡L kãwUi Aÿi ̧wj‡K KZ cÖKv‡i mvRv‡bv hvq? (M) kãwU †_‡K cÖwZ ev‡i wZbwU K‡i eY© wb‡q KZ ̧wj kã MVb Kiv hvq? mgvavb: (K) †gvevBj b¤^i ̧wj 11 A‡1⁄4i Ges wbw`©ó A1⁄4 3 wU|  cwieZ©bkxj A1⁄4 (11 – 3) ev 8 wU| Avevi, †gvU MvwYwZK A1⁄4 10 wU|  †gvU ms‡hvM †`Iqv hv‡e 108 wU| (L) PERMUTATION kãwU‡Z †gvU eY© Av‡Q 11 wU| G‡`i g‡a ̈ ̄^ieY© 5 wU Ges e ̈ÄbeY© 6 wU| GLv‡b 5 wU †Rvo ̄’v‡b 5 wU ̄^ieY© Øviv c~iY Kivi Dcvq 5! Aewkó 6 wU ̄’vb 6 wU e ̈ÄbeY© Øviv c~iY Kivi Dcvq 6! 2!  wb‡Y©q mvRv‡bv msL ̈v = 5!  6! 2! = 43200 (M) PERMUTATION kãwU‡Z †gvU Aÿi Av‡Q 11 wU| G‡`i g‡a ̈ 2 wU T Av‡Q Ges evwK me wfbœ wfbœ| cÖwZevi wZbwU eY© wb‡q mvRv‡bvi Dcvq: (i) `ywU T I GKwU wfbœ n‡j = 9C1  3! 2! (ii) wZbwU wfbœ eY© n‡j = 10C3  3!  wb‡Y©q kã MVb msL ̈v = 9C1  3! 2! + 10C3  3! = 27 + 720 = 747 10| Av`gRx K ̈v›Ub‡g›U K‡j‡R MATHEMATICS Gi wkÿK msL ̈v 6 Rb Ges PHYSICS Gi wkÿK msL ̈v 5 Rb| (K) 2n + 1Pn + 1 : 2nPa + 2 = 5 : 2 n‡j n Gi gvb wbY©q Ki| (L) MATHEMATICS Gi wkÿK‡`i msL ̈vMwiôZv w`‡q 8 Rb wkÿ‡Ki GKwU KwgwU KZ iK‡g MVb Kiv hvq? (M) DÏxc‡Ki 1g Bs‡iwR kãwU n‡Z 6 wU K‡i eY© KZ iK‡g evQvB Kiv hvq? mgvavb: (K) 2n + 1Pn + 1 : 2nPa + 2 = 5 : 2  2n + 1Pn + 1 2nPn + 2 = 5 2  (2n + 1)! {(2n + 1) – (n + 1)}! (2n)! {2n – (n + 2)}! = 5 2  (2n + 1)! n! (2n)! (n – 2)! = 5 2  (2n + 1)! n!  (n – 2)! (2n)! = 5 2  (2n + 1) . (2n)! n(n – 1) (n – 2)!  (n – 2)! (2n)! = 5 2  2n + 1 n 2 – n = 5 2  5n2 – 5n = 4n + 2  5n2 – 9n – 2 = 0  5n2 – 10n + n – 2 = 0  5n(n – 2) + 1(n – 2) = 0  (n – 2) (5n + 1) = 0  n = 2    ⸪ n   – 1 5