Home
Class 12
MATHS
Find the number of all three elements su...

Find the number of all three elements subsets of the set`{a_1, a_2, a_3, a_n}` which contain `a_3dot`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of finding the number of all three-element subsets of the set {a_1, a_2, a_3, ..., a_n} that contain the element a_3, we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Fixed Element**: Since the subset must contain the element a_3, we can consider a_3 as a fixed element in our subset. 2. **Count Remaining Elements**: After fixing a_3, we are left with the remaining elements of the set, which are {a_1, a_2, a_4, a_5, ..., a_n}. This means we have (n - 1) elements left to choose from. 3. **Determine the Number of Elements to Choose**: Since we need a total of three elements in our subset and one of them is already chosen (a_3), we need to select 2 more elements from the remaining (n - 1) elements. 4. **Use the Combination Formula**: The number of ways to choose 2 elements from (n - 1) elements can be calculated using the combination formula: \[ \binom{n-1}{2} \] where \(\binom{n-1}{2}\) is the number of combinations of (n - 1) elements taken 2 at a time. 5. **Final Answer**: Therefore, the number of all three-element subsets of the set {a_1, a_2, a_3, ..., a_n} that include the element a_3 is given by: \[ \binom{n-1}{2} \]

To solve the problem of finding the number of all three-element subsets of the set {a_1, a_2, a_3, ..., a_n} that contain the element a_3, we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Fixed Element**: Since the subset must contain the element a_3, we can consider a_3 as a fixed element in our subset. 2. **Count Remaining Elements**: After fixing a_3, we are left with the remaining elements of the set, which are {a_1, a_2, a_4, a_5, ..., a_n}. This means we have (n - 1) elements left to choose from. ...
Promotional Banner

Similar Questions

Explore conceptually related problems

The number of all three element subsets of the set {a_(1),a_(2),a_(3)......a_(n)} which contain a_(3) , is

The number of all subsets of a set containing 2n+1 elements which contains more than n elements is

A sequence of number is represented as a_1, a_2, a_3,….a_n . Each number in the sequence (except the first and the last) is the mean of the first two adjacent numbers in the sequece. If a_(1) = 1 and a_5 = 3 , what is hte value of a_(3) ?

Find the number of ways of arranging 15 students A_1,A_2,........A_15 in a row such that (i) A_2 , must be seated after A_1 and A_2 , must come after A_2 (ii) neither A_2 nor A_3 seated brfore A_1

Let a_1=0 and a_1,a_2,a_3 …. , a_n be real numbers such that |a_i|=|a_(i-1) + 1| for all I then the A.M. Of the number a_1,a_2 ,a_3 …., a_n has the value A where : (a) A lt -1/2 (b) A lt -1 (c) A ge -1/2 (d) A=-2

The terms a_1, a_2, a_3 from an arithmetic sequence whose sum s 18. The terms a_1+1,a_2, a_3,+2, in that order, form a geometric sequence. Then the absolute value of the sum of all possible common difference of the A.P. is ________.

If (b_2-b_1)(b_3-b_1)+(a_2-a_1)(a_3-a_1)=0 , then prove that the circumcenter of the triangle having vertices (a_1,b_1),(a_2,b_2) and (a_3,b_3) is ((a_2+a_3)/(2),(b_2+b_3)/(2)) .

Find the indicated terms in each of the following sequences whose nth terms are: a_n=5_n-4; a_(12) and a_(15) a_n=(3n-2)/(4n+5); a_7 and a_8 a_n=n(n-1); a_5 and a_8 a_n=(n-1)(2-n)(3+n); a_1,a_2,a_3 a_n=(-1)^nn ; a_3,a_5, a_8

If a_1,a_2,a_3,.....,a_(n+1) be (n+1) different prime numbers, then the number of different factors (other than1) of a_1^m.a_2.a_3...a_(n+1) , is

Arithmetic mean a, geometric mean G and Harmonic mean H to n positive numbers a_1,a_2,a_3,…..,a_n are given by A=(a_1+a_2+……………+a_n)/n, G=(a_1 a_2 a_n)^(1/2) and G= n/(1/H_1+1/H_2+………+1/H_n) There is a relation in A, G and H given by AgeGgeH equality holds if and only if a_1=a_2=..............=a_n On the basis of above information answer the following question If a_rgt0 or r=1,2,.......6 and a_1+a_2+.....+a_6=3, M=(a_1+a_2)(a_3+a_4)(a_5+a_6) . Then the set of all possible values of M is (A) (0,3] (B) (0,2] (C) (0,1] (D) none of these