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There are unlimited number of coupons each bearing one of the letters of A,B,C .In how many ways  "m" coupons can be selected such that the word BAC cannot be spelt. [answer is 3*(2^m - 1) ]
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# Other Related Questions on Algebra

how to find number of positive integral solution of x+y+z=20

Min values of x,y and z will be 1 so remaining 17 have to be distributed over these numbers Just divide 17 in three parts using 2 partitions which can be done by 19C2 ways as 2 paritions...

 Harsh Patodia one month ago
If for an AP A1,A2,A3....an. A1+A3+A5=-12 and A1*A2*A3=8. Find the value of A2+A4+A6.

Shorter method Let d be the common difference . (A3-2d)+ A3 + (A3+2d) = -12 and A3 = -4 (A3-2d)* (A3-d) *(A3) = 8 Solve for d Now A2+A4+A6 = (A3-d) + (A3+d) + (A3+2d). Substitute values for ...

 Ajay 2 months ago

Let d be the common difference . A1+ (A1+2d) + (A1+4d) = -12 and A1*(A1+2d) *(A1+4d) = 8 and solve equations for A1 and d Now A2+A4+A6 = (A1+d) + (A1+3d) + (A1+5d). Substitute values for A1 ...

 Ajay 2 months ago
solution set of inequility (3 x -4 x ) ln(x+2) ÷ x 2 -3x-4

Where is inequality in this question? i can see only expression. Kindly recheck and repost the question.

 Harsh Patodia one month ago
solve this vectors of OA=I+J+K ; AB=3I-2J+K; BC=I+2J-2K and CD =2I+J+3K FIND OD VECTOR

p { margin-bottom: 0.25cm; line-height: 120%; } GIVEN VECTORS OA=I+J+K;AB=3I-2J+K;BC=I+2J-2K AND CD=2I+J+3K WE KNOW THAT VECTOR AB =OB-OA 3I-2J+K= OB-( I+J+K) OB=4I-J+2K SIMILARLY BC =OC-OB ...

 PASUPULETI GURU MAHESH 7 months ago
Find the derivative of f(x) = x 3 -x using the definition of derivative ?

The given function is f(x) = x 3 -x Now we calculate the difference quotient f(x+h) – f(x) / h = [(x+h) 3 - (x+h)] – [x 3 -x] / h 3x 2 +3xh+h 2 -1 lim h-0 3x 2 +3xh+h 2 -1 = 3x 2 - 1

 Mounik 7 months ago
I want an proper algorithm to solve the problem?...................................

solution in the figure

 Riddhish Bhalodia 5 months ago
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