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KVPY Sample Paper KVPY Stream-SX Model Paper-18

  • question_answer
    Let \[{{n}_{1}}<{{n}_{2}}<{{n}_{3}}<{{n}_{4}}<{{n}_{5}}\]be positive integers such that \[{{n}_{1}}+{{n}_{2}}+{{n}_{3}}+{{n}_{4}}+{{n}_{5}}=20.\] Then, the number of such distinct arrangement \[({{n}_{1}},{{n}_{2}},{{n}_{3}},{{n}_{4}},{{n}_{5}})\] is

    A) 7

    B) 6

    C) 8

    D) 5

    Correct Answer: A

    Solution :

    We have, \[{{n}_{1}},\]\[{{n}_{2}},\]\[{{n}_{3}},\]\[{{n}_{4}},\]\[{{n}_{5}}\] are positive integer and  \[{{n}_{5}}>{{n}_{4}}>{{n}_{3}}>{{n}_{2}}>{{n}_{1}}\]
    \[\therefore \]\[{{n}_{1}}\ge 1,\]\[{{n}_{2}}\ge 2,\]\[{{n}_{3}}\ge 3,\]\[{{n}_{4}}\ge 4,\]\[{{n}_{5}}\ge 5\]
    Let \[{{n}_{1}}-1={{x}_{1}}\ge 0,\]\[{{n}_{2}}-2={{x}_{2}}\ge 0\]
    \[{{x}_{5}}-5={{x}_{5}};{{x}_{5}}\ge 0\]
    \[\therefore \]\[{{x}_{1}}+1+{{x}_{2}}+2+...+{{x}_{5}}+5=20\]
    \[\Rightarrow \]\[{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}+{{x}_{5}}=5\]
    Now, \[{{x}_{1}}\le {{x}_{2}}\le {{x}_{3}}\le {{x}_{4}}\le {{x}_{5}}\]
    \[{{x}_{1}}\] \[{{x}_{2}}\] \[{{x}_{3}}\] \[{{x}_{4}}\] \[{{x}_{5}}\]
    0 0 0 0 5
    0 0 0 1 4
    0 0 0 2 3
    0 0 1 1 3
    0 0 1 2 2
    0 1 1 1 2
    1 1 1 1 1
    So, 7 possible cases will be there.


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