સમીકરણ x1 + x2 + x3 + x4 = 31, x1, x2, x3, x4 ∈ W ના 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.
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Answer
    • 32.
    15 સમાન વસ્તુઓને ત્રણ પેટીમાં કુલ કેટલી રીતે મૂકી શકાય ? (દરેક પેટીમાં વધુમાં વધુ 15 વતુ આવી શકે.)
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    Answer
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    33. 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

    A.

    open parentheses table row bold 34 row bold 3 end table close parentheses

    Tips: -

    માંગેલ ઉકેલોની સંખ્યા = (1 + x + x2 + ... + x31)4 ના વિસ્તરણમાં x31 નો સહગુણક
     
                             = fraction numerator bold left parenthesis bold 1 bold minus bold x bold right parenthesis to the power of bold 4 bold space bold left parenthesis bold 1 bold space bold plus bold space bold x bold space bold plus bold space bold x to the power of bold 2 bold space bold plus bold space bold. bold. bold. bold space bold plus bold space bold x to the power of bold 31 bold right parenthesis to the power of bold 4 over denominator bold left parenthesis bold 1 bold minus bold x bold right parenthesis to the power of bold 4 end fraction માં x31 નો સહગુણક
     
                             = (1 - x32)4 (1-x)-4 માં x31 નો સહગુણક 

     
                             = (1-x)-4 માં x31 નો સહગુણક 


                              = open parentheses table row cell bold 31 bold plus bold 4 bold minus bold 1 end cell row bold 31 end table close parentheses              (1-x)-n માં xr નો સહગુણકbold space bold equals bold space open parentheses table row cell bold n bold plus bold r bold minus bold 1 end cell row bold r end table close parentheses

                              = bold equals bold space open parentheses table row bold 34 row bold 31 end table close parentheses bold equals bold space open parentheses table row bold 34 row bold 3 end table close parentheses
    Answer
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    34. શબ્દ ASSASSINATION શબ્દના અક્ષરોમાંથી 4 અક્ષરો કુલ કેટલી રીતે પસંદ કરી શકાય ?
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    35. 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
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    Answer
    • 36.
    સમીકરણ x + y + z = 20, x ≥ 1, y ≥ 2, z ≥ 3, x, y, z ∈ Wના ઉકેલોની સંખ્યા ...... છે.
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    Answer
    • 37. અસંખ્ય લાલ, પીળા અને વાદળી રંગના દડામાંથી 12 દડા પસંદ કરવાના પ્રકાર ....... છે. 
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    Answer
    • 38. અસમતા x + y + z ≤ 30,  ના અનૃપ પૂર્ણાંક ઉકેલોની સંખ્યા ....... છે.
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    39. 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
    • 40. 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
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