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Question: If one quarter of all the subsets containing three elements of the integers 1, 2, 3,……..m contain th...

If one quarter of all the subsets containing three elements of the integers 1, 2, 3,……..m contain the integer 5, then m is equal to
(1) 12
(2) 10
(3) 14
(4) 11

Explanation

Solution

Here first we will find the number of subsets of the integers 1, 2, 3,……..m containing three elements which contain the integer 5. Then we will form an equation according to the given data and solve for the value of m to get the required answer.

Complete step-by-step answer:
It is given that:-
One quarter of all the subsets containing three elements of the integers 1, 2, 3,……..m contain the integer 5.
Now the subsets of the integers containing three elements are given by:- mC3^m{C_3}
Also, number of subsets of integers containing three elements in which one element is 5 given by:- m−1C2^{m - 1}{C_2}
Now it is given that, 1/4th of the all the subsets containing three elements of the integers 1, 2, 3,……..m contain the integer 5.
This implies,
14(mC3)=m−1C2\dfrac{1}{4}\left( {^m{C_3}} \right){ = ^{m - 1}}{C_2}
Now we will use the following formula:-
nCr=n!r!(n−r)!^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}
Applying this formula we get:-
14(m!3!(m−3)!)=(m−1)!2!(m−1−2)!\dfrac{1}{4}\left( {\dfrac{{m!}}{{3!\left( {m - 3} \right)!}}} \right) = \dfrac{{\left( {m - 1} \right)!}}{{2!\left( {m - 1 - 2} \right)!}}
Solving it further we get:-
14(m×(m−1)!3×2!(m−3)!)=(m−1)!2!(m−3)!\dfrac{1}{4}\left( {\dfrac{{m \times \left( {m - 1} \right)!}}{{3 \times 2!\left( {m - 3} \right)!}}} \right) = \dfrac{{\left( {m - 1} \right)!}}{{2!\left( {m - 3} \right)!}}
Cancelling the terms we get:-
14(m3)=1\dfrac{1}{4}\left( {\dfrac{m}{3}} \right) = 1
Now solving for m we get:-

m12=1 ⇒m=12  \dfrac{m}{{12}} = 1 \\\ \Rightarrow m = 12 \\\

Therefore, the value of m is 12.

Hence, option (1) is the correct option.

Note: Students can also solve the given problem by finding the difference between the number of subsets of integers containing three elements and the number of three element subsets of integers which do not contain element 5.
Number of subsets of integers containing three elements in which one element is 5 = number of subsets of integers containing three elements - the number of three element subsets of integers which do not contain element 5
Hence, we get:
Number of subsets of integers containing three elements in which one element is 5=mC3−m−1C3{\text{Number of subsets of integers containing three elements in which one element is 5}}{ = ^m}{C_3}{ - ^{m - 1}}{C_3}