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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

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2^{m} - 1)

2^{m - 1} - 1)

{3 (2}^{m} + 1)

None of these

answer is A.

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Detailed Solution

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)

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