If there are ten positive real numbers n1 < n2 < n3 … < n10, how many triplets of these numbers (n1, n2, n3), (n2, n3, n4) … can be generated such that in each triplet the first number is always less than the second number, and the second number is always less than the third number?  Choose the correct answer:	A) 45	B) 90	C) 120	D) 180	E) 150
Three numbers can be selected and arranged out of ten numbers in 10P3 ways=10!/7!=10*9*8. Now this arrangement is restricted to a given condition that first number is always less than the second number, and the second number is always less than the third number. Hence three numbers can be arranged among themselves in 3! ways.   Required number of arrangements=(10*9*8)/(3*2)= 
<gadget id="calculator">(10*9*8)/(3*2)</gadget>
<output>120</output>
120   ANS:120
<result>C</result>
If there are ten positive real numbers n1 < n2 < n3 … < n10, how many triplets of these numbers (n1, n2, n3), (n2, n3, n4) … can be generated such that in each triplet the first number is always less than the second number, and the second number is always less than the third number?