(1) से (10) तक की संख्याओं में से (3) संख्याएँ ऐसी चुननी हैं कि कोई दो क्रमागत न हों। कुल चयन कितने हैं?
From the numbers (1) to (10), (3) numbers are to be selected so that no two are consecutive. How many selections are possible?
Explanation opens after your attempt
Correct Answer
A. (56)
Concept
The no-consecutive formula gives \( \binom{10-3+1}{3}=\binom{8}{3}=56 \). In such questions, the gap method is very fast.
Why this answer is correct
The correct answer is A. (56). The no-consecutive formula gives \( \binom{10-3+1}{3}=\binom{8}{3}=56 \). In such questions, the gap method is very fast.
Exam Tip
क्रमागत न होने का सूत्र \( \binom{10-3+1}{3}=\binom{8}{3}=56 \) देता है। ऐसे प्रश्नों में खाली स्थान विधि बहुत तेज होती है।
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