Hai guyz in the pic solve the 18th question.The final answer is given in the pic.I don't understand why 2 comes in the denominator.Plz solve with steps
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kvnmurty:
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11 books are different. But 4 packets A, B, C, and D are identically labeled from outside.
Ways of packing A: 11C3.
Ways of packing B: 8C3.
Ways of packing C: 5C3.
Ways of packing D: 2C2.
Their product = 11! /[(3!)^3 2!].
After packing Interchanging among A B and C doesn't give new way. But these interchangings are counted in the product calculated above. Divide by 3!.
Answer = 11!/[ (3!)^4 * 2!] = 15400
Ways of packing A: 11C3.
Ways of packing B: 8C3.
Ways of packing C: 5C3.
Ways of packing D: 2C2.
Their product = 11! /[(3!)^3 2!].
After packing Interchanging among A B and C doesn't give new way. But these interchangings are counted in the product calculated above. Divide by 3!.
Answer = 11!/[ (3!)^4 * 2!] = 15400
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