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Question: The total number of ways in which three distinct numbers in AP can be selected from the set {1, 2, 3...

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

Explanation

Solution

In order to solve this question, we should have some knowledge of combination, which is expressed mathematically as nCr=n!r!(n−r)!^{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−1Cr−1^{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 α,β and γ\alpha ,\beta \text{ and }\gamma . So, we can say the sum of the number of terms before α\alpha , number of terms between α and β\alpha \text{ and }\beta , number of terms between β and γ\beta \text{ and }\gamma and the number of terms after γ\gamma and 3 (because of α,β,γ\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 α\alpha , y = number of terms between α and β,β and γ\alpha \text{ and }\beta ,\beta \text{ and }\gamma and z = number of terms after γ\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≥0x+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−1Cr+1^{n+r-1}{{C}_{r+1}}.
So, for y = 0,
x+z=21−2(0)x+z=21-2\left( 0 \right)
x+z=21x+z=21

Then the number of ways for choosing x and z is
21+2−1C2−1= 22C1^{21+2-1}{{C}_{2-1}}={{\text{ }}^{22}}{{C}_{1}}
For y = 1,
x+z=21−2(1)x+z=21-2\left( 1 \right)
x+z=19x+z=19

Then the number of ways for choosing x and z is
19+2−1C2−1= 20C1^{19+2-1}{{C}_{2-1}}={{\text{ }}^{20}}{{C}_{1}}
For y = 2,
x+z=21−2(2)x+z=21-2\left( 2 \right)
x+z=17x+z=17

Then the number of ways for choosing x and z is
17+2−1C2−1= 18C1^{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≥021-2y\ge 0 gives y≤10.5y\le 10.5. Hence, the maximum possible value of y is 10.
For y = 10,
x+z=21−2(10)x+z=21-2\left( 10 \right)
x+z=1x+z=1

Then the number of ways for choosing x and z is
1+2−1C2−1= 2C1^{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
22C1+ 20C1+ 18C1.....+ 2C1^{22}{{C}_{1}}+{{\text{ }}^{20}}{{C}_{1}}+{{\text{ }}^{18}}{{C}_{1}}.....+{{\text{ }}^{2}}{{C}_{1}}
And we know that,
nCr=n!r!(n−r)!^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}
So, for r = 1, we can say that,
nC1=n!1!(n−1)!=n^{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=n2(a+l)S=\dfrac{n}{2}\left( a+l \right)
Therefore, we can say from the formula l = a + (n – 1)d, we get,
2=22+(n−1)(−2)2=22+\left( n-1 \right)\left( -2 \right)
2−22=(−2)(n−1)2-22=\left( -2 \right)\left( n-1 \right)
−20−2=n−1\dfrac{-20}{-2}=n-1
10=n−110=n-1
n=11n=11

And therefore, we can write the sum as,
22+20+18+.....+2=112(22+2)22+20+18+.....+2=\dfrac{11}{2}\left( 22+2 \right)
22+20+18+.....+2=112×2422+20+18+.....+2=\dfrac{11}{2}\times 24
22+20+18+.....+2=11×12=13222+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=n2(a+l)S=\dfrac{n}{2}\left( a+l \right)
Also, we have to remember for x1+x2+x3+....+xr=n{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+....+{{x}_{r}}=n the possible number of ways of choosing x1,x2,x3,....xr=n{{x}_{1}},{{x}_{2}},{{x}_{3}},....{{x}_{r}}=n are n+r−1Cr−1^{n+r-1}{{C}_{r-1}}. By using them, we can solve the question.