સમીકરણ x + y + z = 20, x ≥ 1, y ≥ 2, z ≥ 3, x, y, z ∈ Wના ઉકેલોની સંખ્યા ...... છે. from Mathematics દ્વિપદી પ્રમેય
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Gujarati JEE Mathematics : દ્વિપદી પ્રમેય

Multiple Choice Questions

31. (1 + x + x2 + x3 + x4)3 ના વિસ્તરણમાં x5 નો સહગુણક ...... છે.
  • 21
  • 24
  • 18
  • 15
Answer
    • 32. અસમતા x + y + z ≤ 30,  ના અનૃપ પૂર્ણાંક ઉકેલોની સંખ્યા ....... છે.
    • 5646
    • 5456
    • 5465
    • 5654
    Answer
    • 33. જો a અને b ધન પૂર્ણાંક હોય, તો open parentheses table row bold a row bold b end table close parentheses bold space bold plus bold space open parentheses table row bold a row cell bold r bold minus bold 1 end cell end table close parentheses bold space open parentheses table row bold b row bold 1 end table close parentheses bold space bold plus bold space open parentheses table row bold a row cell bold r bold minus bold 2 end cell end table close parentheses bold space open parentheses table row bold b row bold 2 end table close parentheses bold space bold plus bold space bold. bold. bold. bold space bold plus bold space open parentheses table row bold b row bold r end table close parentheses bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold. bold space 
    • open parentheses table row cell bold a bold plus bold b end cell row bold r end table close parentheses
    • open parentheses table row cell bold a bold plus bold b end cell row cell bold r bold plus bold 1 end cell end table close parentheses
    • fraction numerator bold left parenthesis bold a bold plus bold b bold right parenthesis bold factorial over denominator bold r bold factorial end fraction
    • fraction numerator bold a bold factorial bold b bold factorial over denominator bold r bold factorial end fraction
    Answer
    • 34.
    15 સમાન વસ્તુઓને ત્રણ પેટીમાં કુલ કેટલી રીતે મૂકી શકાય ? (દરેક પેટીમાં વધુમાં વધુ 15 વતુ આવી શકે.)
    • 136
    • 315
    • 15P3
    • 91
    Answer
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    35. અસંખ્ય લાલ, પીળા અને વાદળી રંગના દડામાંથી 12 દડા પસંદ કરવાના પ્રકાર ....... છે. 
    • 55
    • 91
    • 105
    • અસંખ્ય
    Answer
    • 36. જો bold c subscript bold r bold space bold equals open parentheses table row bold 30 row bold r end table close parentheses bold space bold ત ો bold space open parentheses bold c subscript bold 1 over bold c subscript bold 0 close parentheses to the power of bold 2 bold space bold plus bold space bold 8 bold space open parentheses bold c subscript bold 2 over bold c subscript bold 1 close parentheses to the power of bold 2 bold space bold plus bold space bold. bold. bold. bold space bold plus bold space bold 30 to the power of bold 3 bold space open parentheses bold c subscript bold 30 over bold c subscript bold 29 close parentheses to the power of bold 2 bold space bold equals bold space bold. bold. bold. bold. bold space bold. bold space
    • 75880
    • 76080
    • 76880
    • 76808
    Answer
    • 37. bold 3 over bold 4 bold plus open parentheses bold 3 over bold 4 bold cross times bold 5 over bold 8 close parentheses bold space bold plus bold space open parentheses bold 3 over bold 4 bold cross times bold 5 over bold 8 bold cross times bold 7 over bold 12 close parentheses bold space bold plus bold space bold. bold. bold. bold space અનંત પદ= .......
    • bold 2 bold space bold times bold space square root of bold 2 bold space bold plus bold space bold 1
    • square root of bold 2 bold space bold minus bold space bold 1
    • bold 2 bold space bold times bold space square root of bold 2 bold space bold minus bold space bold 1
    • bold 2 bold space bold times bold space square root of bold 2
    Answer
    • 38. શબ્દ ASSASSINATION શબ્દના અક્ષરોમાંથી 4 અક્ષરો કુલ કેટલી રીતે પસંદ કરી શકાય ?
    • 80
    • 71
    • 72
    • 60
    Answer
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    39.
    સમીકરણ x + y + z = 20, x ≥ 1, y ≥ 2, z ≥ 3, x, y, z ∈ Wના ઉકેલોની સંખ્યા ...... છે.
    • 124
    • 136
    • 210
    • 120

    D.

    120

    Tips: -

    માંગેલ ઉકેલોની સંખ્યા 

    = (x1 + x2 + ... + x20) (x2 + x3 + ... + x20) (x3 + x4 + ... + x20) માં x20 નો સહગુણક
     
    = x6 (1 + x + ... + x19) (1 + x + ... + x18) (1 + x + ... + x17) માં x20 નો સહગુણક
     
    = (1 + x + ... + x19) (1 + x + ... + x18) ( 1 + x + ... + x17) માં x14 નો સહગુણક
     
    bold equals bold space open parentheses fraction numerator bold 1 bold minus bold x to the power of bold 20 over denominator bold 1 bold minus bold x end fraction close parentheses bold space open parentheses fraction numerator bold 1 bold minus bold x to the power of bold 19 over denominator bold 1 bold minus bold x end fraction close parentheses bold space open parentheses fraction numerator bold 1 bold minus bold x to the power of bold 18 over denominator bold 1 bold minus bold x end fraction close parentheses માં x14 નો સહગુણક
     
    = (1-x20) (1-x19) (1-x18) (1-x)-3 માં x14 નો સહગુણક
     
    = (1-x)-3 માં x14 નો સહગુણક
    bold equals bold space open parentheses table row cell bold 3 bold plus bold 14 bold minus bold 1 end cell row bold 14 end table close parentheses bold space bold equals bold space open parentheses table row bold 16 row bold 14 end table close parentheses bold space bold equals bold space open parentheses table row bold 16 row bold 2 end table close parentheses bold space bold equals bold space fraction numerator bold 16 bold cross times bold 15 over denominator bold 2 end fraction bold space bold equals bold space bold 120
    Answer
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    40. સમીકરણ x1 + x2 + x3 + x4 = 31, x1, x2, x3, x4 ∈ W ના ઉકેલોની સંખ્યા ........ છે.
    • open parentheses table row bold 34 row bold 3 end table close parentheses
    • open parentheses table row bold 35 row bold 4 end table close parentheses
    • open parentheses table row bold 33 row bold 2 end table close parentheses
    • open parentheses table row bold 34 row bold 4 end table close parentheses
    Answer
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