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From the numbers (1) to (10), (3) numbers are to be selected so that no two are consecutive. How many selections are possible?(1) से (10) तक की संख्याओं में से (3) संख्याएँ ऐसी चुननी हैं कि कोई दो क्रमागत न हों। कुल चयन कितने हैं?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: (56)(56)
The no-consecutive formula gives ( \binom{10-3+1}{3}=\binom{8}{3}=56 ). In such questions, the gap method is very fast.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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