Question

The number of ways in which we can select 4 numbers from 1 to 30 so as to exclude every selection of four consecutive numbers is
a) 27378
b) 27405
c) 27504
d) 27387

Answer 1

Hint: In this question, we have to find out the number of ways in which we can select 4 numbers from 1 to 30 such that four consecutive numbers are not selected. Therefore, as the ordering of the selection of the numbers is not important, i.e. the selection (1,5,7,8) will be the same as (1,7,8,5) and so on, we can find the total number of possible selection as $^{30}{{C}_{4}}=\dfrac{30!}{4!(30-4)!}$ and then subtract the selections which consist of 4 consecutive numbers which will be 27 in number (1,2,3,4), (2,3,4,5)…(27,28,29,30). Thus, we can get the required answer as $^{30}{{C}_{4}}-27$ and then match this value to the given options to obtain the required answer.

Complete step by step solution:
In this question, we are asked to find the number of ways in which we can select 4 numbers from 1 to 30 such that four consecutive numbers are not selected. Thus, we should find the total number of ways in which we can select 4 numbers from 1 to 30.
We know that the total number of ways of selecting r numbers from n numbers where the ordering of selection is not important is given by
$^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}................................(1.1)$
where $x!=1\times 2\times ...(x-1)\times x$ .
Here, there are a total of 30 numbers. Therefore, taking n=30 and r=4 in equation (1.1), we get
Total number of possible selections=
 $\begin{align}
  & ^{30}{{C}_{4}}=\dfrac{30!}{4!\left( 30-4 \right)!}=\dfrac{30\times 29\times 28\times 27\times 26\times ...\times 1}{4\times 3\times 2\times 1\times 26\times 25\times ...\times 1} \\
 & =\dfrac{30\times 29\times 28\times 27}{4\times 3\times 2\times 1}=27405..................(1.2) \\
\end{align}$
However, it is given that we have to exclude all the selection of 4 consecutive numbers. However, as the ordering is not important, the only such selections possible are
(1,2,3,4), (2,3,4,5)…(27,28,29,30)
Thus, there are 27 such selections. Therefore, we should subtract 27 from (1.2) to obtain our required answer.
Thus, the required answer will be 27405-27=27378 which matches option (a). Thus, option (a) is the correct answer to this question.

Note: We should note that in the entire solution, we have not assumed that the ordering of the 4 numbers in a selection is not important. However, if in the question, it were given that the ordering is important, then we should have used permutations $^{30}{{P}_{4}}=\dfrac{30!}{(30-4)!}$ instead of $^{30}{{C}_{4}}$ in equation (1.2). Also, we should take care of the exact selections that should be omitted from the selections to obtain the required answer.