Question

30 children are forming a ring number of ways in which 3 of them can be selected, no two of whom are next to each other, is:
A. $^{26}{C_{24}}$
B. ${10.^{27}}{C_{24}}$
C. $^{27}{C_{25}}$
D. ${10.^{26}}{C_{24}}$

Answer 1

We will first find the total number of ways in which we can select 3 children from thirty using combination formula. Now, we will subtract the number of possibilities where 2 or 3 children can be together in it.

Complete step-by-step solution:
We know that if we need to select r objects from n objects, we will have $^n{C_r}$ possible ways to do the same.
Therefore, to choose 3 children from 30, we will have $^{30}{C_3}$ possible ways to do so.
Now, we just need to subtract the number of possibilities where 2 or 3 children can be sitting together.
Now, let us name the students by their place only.
Let the students be 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29 and 30.
Now, we need to subtract the number of ways the students can be seated together.
Now, let us first draw the seating of children.

Here, the yellow circles are the representation of seats for children.
Now, let us make pairs of two children at a time who are sitting adjacent to each other.
The pairs possible will be (1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7), (7, 8), (8, 9), (9, 10), (10, 11), (11, 12), (12, 13), (13, 14), (14, 15), (15, 16), (16, 17), (17, 18), (18, 19), (19, 20), (20, 21), (21, 22), (22, 23), (23, 24), (24, 25), (25, 26), (26, 27), (27, 28), (28, 29), (29, 30) and (30, 1).
Now, these are 30 pairs. Let us pick one pair at once.
Case I:
Let us first pick (1, 2) in the start. Now, the third children can be any among the rest 28. Hence, the case I has 28 possibilities.
Case II:
Let us now pick (2, 3). Now, the third children can be any among the rest 28. But, it cannot be 1 because 1, 2, 3 are covered in case I only. Hence, case II has 27 possibilities.
Going on like this, we will get 27 possibilities till we reach (29, 30).
Case XXX:
We picked (30, 1). Now, we cannot pick 28 or 2 any because those are covered in previous cases. Hence, this case has 26 possibilities.
Therefore, we need to eradicate the possibilities: 28 + 27 + 27 +……..upto 28 times + 26.
Calculating this, we will get: $28 + (28 \times 27) + 26 = 810$.
Now, we need to subtract this from total possible ways.
Hence, the required answer is: $^{30}{C_3} - 810$.
We know that $^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$.
Therefore, we get:- $^{30}{C_3} - 810 = \dfrac{{30!}}{{3!(27)!}} - 810$.
Thus, we get: $\dfrac{{30 \times 29 \times 28}}{{3 \times 2}} - 810 = 3250$.
Now, let us look at the options one by one.
We first have $^{26}{C_{24}} = \dfrac{{26 \times 25}}{2} = 325$.
We can clearly see that if we multiply it by 10, we will get 3250.

Hence, the answer is (D).

Note: The students must note that it may be possible that you do not understand the options using the first option only and need to evaluate the other options also.
Permutation and Combination is such an interesting and useful topic. Like in this question, if we would have started writing down the possibilities, we will take an eternity to write them. But, we can directly find the number of possibilities using Permutation and Combination.