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The number of ways to select 2 boys and 3 girls from a group of 5 boys and 4 girls is
A$\binom{5}{2} + \binom{4}{3} = 14$
B$\binom{5}{2} \cdot \binom{4}{3} = 40$
C$\binom{5+4}{2+3} = \binom{9}{5} = 126$
D$5 \cdot 4 = 20$
Answer & Solution
Correct answer: B. $\binom{5}{2} \cdot \binom{4}{3} = 40$
1. Two INDEPENDENT selections: pick 2 boys AND pick 3 girls.
2. Number of ways to choose 2 boys from 5: $\binom{5}{2} = 10$.
3. Number of ways to choose 3 girls from 4: $\binom{4}{3} = 4$.
4. By the MULTIPLICATION principle (both selections must happen): $10 \cdot 4 = 40$.
5. Option A uses ADDITION instead of multiplication (would count cases where boys-only OR girls-only chosen). Option C ignores the gender constraint. Option D is the product of total counts (wrong formula).
_Source: NCERT Class 11 Mathematics, Ch 6, §6.5 + Examples (Combinations of two groups), p. 12._
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