An unlimited number of coupons bearing the letters A, B and C are available, then the number of ways of choosing 10 of these coupons so that they can't used to spell BAC

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TestHub · Accepted Answer

Case: I
When all the selected coupons bear the same letter.

One letter can be selected from three letters in $^{3} c_{1}$ ways

⇒ Total number of ways of choosing 10 coupons bearing the same letter is $^{3} c_{1} \times 1 = 3 \underline{Case}$ : II When selected coupons bear two letters only The number of ways of selecting two letters from 3 is $^{3} c_{2}$

⇒ The number of ways in which selected coupons bear two letters only $=^{3} c_{2} (2^{10} - 2)$

Hence required number of ways $= 3 +^{3} c_{2} (2^{10} - 2) = 3 (2^{10} - 1)$

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