Question

The number of ways of choosing $m$ coupons out of an unlimited number of coupons bearing the lettrs $A,B$ and $C$ so they cannot be used to spell the word $BAC,$ is

• A. $3(2^m-1)$
• B. $3(2^{m-1}-1)$
• C. $3(2^m+1)$
• D. None of these

Answer 1

Answer: (A)

$3(2^m-1)$

The word BAC cannot be spelt if the m selected coupons do not contain at least one of A,B and C. Thus the number of ways of selecting m coupons which at A or B ${=2}^m$

This also includes the cases when all the m coupons are A or all are B.

Thus the number of ways of selecting m coupons which at B or C ${=2}^m$

This also includes the cases when all the m coupons are B or all are C.

Number of ways of selecting m coupons which at A or C ${=2}^m$

This also includes the cases when all the m coupons are A or all are C.

Number of ways of selecting m coupons which are all A${=1}^m$

Number of ways of selecting m coupons which are all B${=1}^m$

Number of ways of selecting m coupons which are all C${=1}^m$

So required number${=2}^m+2^m+2^m-1^m-1^m-1^m=3(2^m-1)$