AP EAMCET · Maths · Permutation Combination
The number of integers between 10 and 10,000 such that in every integer every digit is greater than its immediate preceeding digit, is
- A \(1112\)
- B \(437\)
- C \(246\)
- D \(182\)
Answer & Solution
Correct Answer
(C) \(246\)
Step-by-step Solution
Detailed explanation
\(\text{2-digit numbers: } \binom{9}{2} = \frac{9 \times 8}{2} = 36\) \(\text{3-digit numbers: } \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84\)…
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