Question

The number of ways of choosing \[m\] coupons out of an unlimited number of coupons bearing the letters A, B and C so they cannot be used to spell the word BAC is
A. \[3\left( {{2^m} - 1} \right)\]
B. \[3\left( {{2^{m - 1}} - 1} \right)\]
C. \[3\left( {{2^m} + 1} \right)\]
D.None of these

Answer 1

Hint: Here, we are asked to find the number of ways of choosing \[m\] coupons such that the combination BAC cannot be formed. For this, we need to check what are the cases in which the word BAC cannot be formed. If one of the letters is missing from B, A or C we cannot spell the word BAC. Use this logic to find the required number of ways.

Complete step-by-step answer:
Given, number of coupons is \[m\]
The letters present in the coupons are A, B and C
For choosing coupons such that after choosing the word BAC cannot be spelt, one of the letters from B, A and C must be missing, that is only two of the given letters must be present.
There can be three cases for choosing only two letters from A,B and C
Choosing either A or B
We are asked to choose \[m\] coupons, as there are two letters each coupon can have two values. Therefore number of ways of selecting \[m\] coupons will be \[{2^m}\]
 \[W({\text{A or B}}) = {2^m}\] (i)
Choosing either B or C
Similarly, here also we can have two values for each coupon. Therefore number of ways of selecting \[m\] coupons will be \[{2^m}\]
 \[W({\text{B or C}}) = {2^m}\] (ii)
Choosing either A or C
Here also, there are two values for each coupon. Therefore the number of ways of selecting \[m\] coupons will be \[{2^m}\].
 \[W({\text{A or C}}) = {2^m}\] (iii)
There can also be a case where all coupons have the same letters. And in the above cases the coupons with all the same letters are also included, so we will need to subtract that part. As there are three letters so there will be three ways of choosing \[m\] coupons such that all coupons have the same letter. Therefore number of ways of selecting \[m\]
 \[W({\text{same letters}}) = 3\] (iv)
Now, for the total number of ways we need to add all the cases and subtract the case having all coupons with the same letter.
 \[\therefore W({\text{total}}) = W({\text{A or B}}) + W({\text{B or C}}) + W({\text{A or C}}) - W({\text{same letters}})\] (v)
Using equations (i), (ii), (iii) and (iv) in (v), we get
 \[ \Rightarrow W({\text{total}}) = {2^m} + {2^m} + {2^m} - 3\]
 \[ \Rightarrow W({\text{total}}) = 3\left( {{2^m}} \right) - 3\]
 \[ \Rightarrow W({\text{total}}) = 3\left( {{2^m} - 1} \right)\]
Therefore the required answer is \[3\left( {{2^m} - 1} \right)\] number of ways.
Hence, the correct answer is option (A) \[3\left( {{2^m} - 1} \right)\]
So, the correct answer is “Option A”.

Note: In such types of questions, all check carefully the given information and apply all the conditions given in the questions. While finding the number of ways also check whether there is repetition or not, if there is repetition you will need to exclude that part or else you get a value more than the actual value required.