edited by
318 views
0 votes
0 votes

$n_1, n_2, n_3, \dots, n_{10}$ are 10 numbers such that $n_1 > 0$ and the numbers are given in ascending order. How many triplets can be formed using these numbers such that in each triplet, the first number is less than the second number, and the second number is less than the third number?

  1. $109$
  2. $27$
  3. $36$
  4. None of these
edited by

Please log in or register to answer this question.

Related questions

0 votes
0 votes
0 answers
1
go_editor asked May 2, 2016
590 views
The $n$th element of a series is represented as$X_n = (-1)^n x_{n-1}$If $X_0 =x$ and $x>0$, then which of the following is always true?$X_n$ is positive if $n$ is even$X_...
0 votes
0 votes
0 answers
2
0 votes
0 votes
0 answers
3
go_editor asked Mar 2, 2016
276 views
The remainder when $2^{256}$ is divided by $17$ is$7$$13$$11$$1$
0 votes
0 votes
0 answers
4
go_editor asked Mar 2, 2016
289 views
If $\text{U, V, W}$ and $m$ are natural numbers such that $\text{U}^m + \text{V}^m = \text{W}^m$, then which of the following is true?$m < \min\text{(U, V, W)}$$m \max\t...
1 votes
1 votes
0 answers
5
go_editor asked Mar 2, 2016
305 views
For all integers $n>0, \: \: 7^{6n} - 6^{6n}$ is divisible by$13$$128$$549$None of these