Quant Question Of The Day: 149
Counting
In how many ways 3 numbers can be chosen from first 10 natural numbers such that there are at least two numbers between any two selected numbers?
1. 10
2. 15
3. 20
4. 25
Let, a = number of numbers less than the first number selected. b = number of numbers between the first and the second number selected. c= number of numbers between the second and the third number selected. d= number of numbers greater than the third number selected. So, a + b + c + d = 7 But, minimum value of b and c can be 2. So, we need to divide 3 identical things among 4 parts. Hence, answer = 6C3 = 20.
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Quant Question Of The Day: 148
Quant Question Of The Day: 147
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