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NDA 2018
Exam
NDA
Year
2018
Subject
Mathematics
Chapter
Permutation and Combination

NDA 2018 · Mathematics · Permutation and Combination

Where A^ is the complement of A . Let x be the number of integers...

Where A′A^{\prime} is the complement of AA. Let xx be the number of integers lying between 2999 and 8001 which have at least two digits equal. Then x is equal to

Options

  1. A2480
  2. B2481
  3. C2482
  4. D2483

Correct answer

B. 2481

Step-by-step solution

Number of numbers between 2999 and 8001 =8001−2999−1=5001<tableclass="qx−match−listqx−match−table"border="1"cellpadding="6"cellspacing="0"><tr><td>3</td><td></td><td></td><td></td></tr></table>Numberofnumberswithalldigitdistinctandhaving3asstartingdigit=8001-2999-1=5001 <table class="qx-match-list qx-match-table" border="1" cellpadding="6" cellspacing="0"><tr><td>3</td><td></td><td></td><td></td></tr></table> Number of numbers with all digit distinct and having 3 as starting digit=9 \times 8 \times 7=504Numberofnumberswithalldigitdistinctandhaving4asstartingdigit.<br/>Similarlynumberofnumberswithstartingdigit5,6and7respectivelyare504,504and504.TotalnumbersNumber of numbers with all digit distinct and having 4 as starting digit. <br/> Similarly number of numbers with starting digit 5, 6 and 7 respectively are 504, 504 and 504. Total numbers=5 \times 504=2520Hence,requirednumberHence, required number=5001-2520=2481
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