Question

The total number of ways in which three distinct numbers in AP can be selected from the set {1, 2, 3, …. 24} is equal to
(a) 66
(b) 132
(c) 198
(d) None of these

Answer 1

Hint: In order to solve this question, we should have some knowledge of combination, which is expressed mathematically as \[^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]. Also, we have to remember that for r terms, if the sum is n, then the possible number of ways of choosing r terms, integral value is given by \[^{n+r-1}{{C}_{r-1}}\]. By using them, we can solve this question.

Complete step-by-step answer:
In this question, we have been asked to find the number of ways in which three distinct numbers in AP can be selected from the set {1, 2, 3, …..24}. So, to solve this question, let us consider that from 1 to 24, 3 distinct numbers in AP are selected like \[\alpha ,\beta \text{ and }\gamma \]. So, we can say the sum of the number of terms before \[\alpha \], number of terms between \[\alpha \text{ and }\beta \], number of terms between \[\beta \text{ and }\gamma \] and the number of terms after \[\gamma \] and 3 (because of \[\alpha ,\beta ,\gamma \]) will give as 24 because the total number of terms in the set are 24.
So, if we consider x = number of terms before \[\alpha \], y = number of terms between \[\alpha \text{ and }\beta ,\beta \text{ and }\gamma \] and z = number of terms after \[\gamma \], then we can write
x + y + y + z + 3 = 24
x + 2y + z = 21
And we can further write it as
x + z = 21 – 2y
Now, we know that x, y, z can have value either 0 or greater than 0. So, we can write it as,
\[x+z=21-2y,x,y,z\ge 0\]

Now, we will consider the values of y to find the possible combination of x + z. Now, we know that if the sum of r terms gives n, then the possible number of ways of choosing r integral value which will give the sum as n are \[^{n+r-1}{{C}_{r+1}}\].
So, for y = 0,
\[x+z=21-2\left( 0 \right)\]
\[x+z=21\]

Then the number of ways for choosing x and z is
\[^{21+2-1}{{C}_{2-1}}={{\text{ }}^{22}}{{C}_{1}}\]
For y = 1,
\[x+z=21-2\left( 1 \right)\]
\[x+z=19\]

Then the number of ways for choosing x and z is
\[^{19+2-1}{{C}_{2-1}}={{\text{ }}^{20}}{{C}_{1}}\]
For y = 2,
\[x+z=21-2\left( 2 \right)\]
\[x+z=17\]

Then the number of ways for choosing x and z is
\[^{17+2-1}{{C}_{2-1}}={{\text{ }}^{18}}{{C}_{1}}\]
And so on up to y = 10 because x + z will always be greater than 0. So, 21 – 2y should be greater than 0. Therefore, \[21-2y\ge 0\] gives \[y\le 10.5\]. Hence, the maximum possible value of y is 10.
For y = 10,
\[x+z=21-2\left( 10 \right)\]
\[x+z=1\]

Then the number of ways for choosing x and z is
\[^{1+2-1}{{C}_{2-1}}={{\text{ }}^{2}}{{C}_{1}}\]
Hence, we can say that the number of all possible combinations of x + 2y + z = 21 will be the sum of the number of possible combinations for the different values of y. So, we can write
Total number of ways of selecting 3 distinct number in AP from 1 to 24
\[^{22}{{C}_{1}}+{{\text{ }}^{20}}{{C}_{1}}+{{\text{ }}^{18}}{{C}_{1}}.....+{{\text{ }}^{2}}{{C}_{1}}\]
And we know that,
\[^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
So, for r = 1, we can say that,
\[^{n}{{C}_{1}}=\dfrac{n!}{1!\left( n-1 \right)!}=n\]

Therefore, we can write the total number of ways of selecting 3 distinct numbers in AP from 1 to 24
= 22 + 20 + 18 …… + 2

Now, we know that the above expression is forming an arithmetic progression with the first term a = 22 and common difference d = 20 – 22 = – 2 and last term l = 2, we can find the number of terms in AP by using the formula l = a + (n – 1)d and then the sum of the terms by using the formula,
\[S=\dfrac{n}{2}\left( a+l \right)\]
Therefore, we can say from the formula l = a + (n – 1)d, we get,
\[2=22+\left( n-1 \right)\left( -2 \right)\]
\[2-22=\left( -2 \right)\left( n-1 \right)\]
\[\dfrac{-20}{-2}=n-1\]
\[10=n-1\]
\[n=11\]

And therefore, we can write the sum as,
\[22+20+18+.....+2=\dfrac{11}{2}\left( 22+2 \right)\]
\[22+20+18+.....+2=\dfrac{11}{2}\times 24\]
\[22+20+18+.....+2=11\times 12=132\]
Hence, the total number of ways of selecting 3 distinct numbers in AP from 1 to 24 is 132.

Therefore, option (b) is the right answer.

Note: While solving this question, we need to remember that the sum of the n terms of an AP, for the first term = a and the last term = l is given by
\[S=\dfrac{n}{2}\left( a+l \right)\]
Also, we have to remember for \[{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+....+{{x}_{r}}=n\] the possible number of ways of choosing \[{{x}_{1}},{{x}_{2}},{{x}_{3}},....{{x}_{r}}=n\] are \[^{n+r-1}{{C}_{r-1}}\]. By using them, we can solve the question.