A) \[\frac{^{16}{{C}_{2}}}{^{40}{{C}_{5}}}\]
B) \[\frac{^{23}{{C}_{2}}}{^{40}{{C}_{5}}}\]
C) \[\frac{^{16}{{C}_{2}}{{\times }^{23}}{{C}_{2}}}{^{40}{{C}_{5}}}\]
D) None of these
Correct Answer: C
Solution :
[c] : Five numbers can be drawn from 40 numbers in \[^{40}{{C}_{5}}\]ways, therefore total number of cases\[{{=}^{40}}{{C}_{5}}\]. We want that \[{{x}_{3}}=24\]. \[\therefore \]The number of favourable cases are \[^{23}{{C}_{2}}{{\times }^{16}}{{C}_{2}}\] Hence, required probability \[=\frac{^{23}{{C}_{2}}{{\times }^{16}}{{C}_{2}}}{^{40}{{C}_{5}}}\]
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