We have 5 consonants and 4 vowels. We need to form a word with 3 consonants and 3 vowels.
\[ \binom{5}{3} = 10 \]
\[ \binom{4}{3} = 4 \]
\[ 6! = 720 \]
\[ 10 \times 4 \times 720 = 28800 \]
Note: If the question means only "selections" of letters (not arrangements), then the answer is:
\[ \binom{5}{3}\times \binom{4}{3} = 10 \times 4 = 40 \]
Final Answer:
- If "word" = arrangement → \(\; \boxed{28800}\)
- If "word" = selection → \(\; \boxed{40}\) (matches given Option 1)
Solution:
Number of straight lines from \(n\) points (no three collinear) is \(\binom{n}{2}\).
Here, \(n = 15\).
\[ \binom{15}{2} = \frac{15 \times 14}{2} = 105 \]
Adjustment for collinearity:
Out of 15 points, 5 are collinear.
Lines from these 5 points = \(\binom{5}{2} = 10\).
But actually they form only 1 line.
Extra counted = \(10 - 1 = 9\).
Correct total lines:
\[ 105 - 9 = 96 \]
Final Answer: The total number of straight lines formed = 96
Case 1: All persons are distinct
For n distinct persons around a round table (rotations same), arrangements = \((n-1)!\).
Here, \(n=9\). So arrangements = \((9-1)! = 8! = \boxed{40{,}320}\).
Case 2: Persons are identical by nationality
If 4 Indians, 3 Americans, and 2 Britishers are considered identical within their groups, then
Arrangements = \[ \frac{(n-1)!}{4!\,3!\,2!} = \frac{8!}{4!\,3!\,2!} = \boxed{140}. \]
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