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Seven distinguishable balls are placed independently and uniformly at…
Question
Seven distinguishable balls are placed independently and uniformly at random into 4 distinguishable boxes (each of the 4^7 placements equally likely). Let p be the probability that no box is empty and box 1 contains exactly 2 balls. Write p = m/n in lowest terms and find m+n.
✓ Verified answer: 9767checked by our engine — not a guess
Step-by-step solution
We need box 1 to contain exactly 2 balls AND boxes 2,3,4 to each be nonempty (box 1 is already nonempty).
Step 1: Choose which 2 of the 7 balls go in box 1: C(7,2) = 21 ways.
Step 2: Distribute the remaining 5 balls among boxes 2,3,4 so that none of these three is empty: this is the number of surjections of a 5-set onto a 3-set = 3!
* S(5,3) = 6 * 25 = 150.
(Equivalently 3^5 - 3*2^5 + 3 = 243 - 96 + 3 = 150.)
Favourable count = 21 * 150 = 3150.
Total placements = 4^7 = 16384.
p = 3150/16384. gcd = 2 -> 1575/8192. Since 8192 = 2^13 and 1575 is odd, this is in lowest terms.
m+n = 1575 + 8192 = 9767.
Final answer9767
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