<p>In how many ways can 5 identical black balls, 7 identical red balls and 6 identical green balls be arranged in a row so that at least one ball is separated from balls of the same colour?</p>
Step-by-Step Solution
Key Concept: Use complementary counting: subtract the number of arrangements where all balls of the same color are together (treated as blocks) from the total unrestricted arrangements of 18 identical balls of three colors.
<p><strong>Step 1:</strong> Find total arrangements of 5 identical black, 7 identical red, and 6 identical green balls in a row.</p><p>Total = $\frac{18!}{5!\cdot 7!\cdot 6!}$ (arranging 18 objects with repetitions)</p><p><strong>Step 2:</strong> Find arrangements where NO ball is separated (all balls of same color are together).</p><p>Treat all black balls as 1 block, all red balls as 1 block, and all green balls as 1 block. We now arrange 3 distinct blocks.</p><p>Number of such arrangements = $3! = 6$</p><p><strong>Step 3:</strong> Apply complementary counting.</p><p>Arrangements where at least one ball is separated from its color group = Total arrangements - Arrangements with all colors grouped</p><p>$$= \frac{18!}{5!\cdot 7!\cdot 6!} - 6$$</p><p><strong>∴ Answer: </strong>$\frac{18!}{5!\cdot 7!\cdot 6!} - 6$</p>
Correct Answer: \(\frac{18!}{5!7!6!} - 6\)