A) \[^{m+n}{{C}_{m}}/{{n}^{m}}\]
B) \[\frac{n\,(n-1)}{(m+n)\,(m+n-1)}\]
C) \[^{m+n}{{P}_{m}}/{{m}^{n}}\]
D) \[^{m+n}{{P}_{n}}/{{n}^{m}}\]
Correct Answer: B
Solution :
\[m\] rupee coins and \[n\] ten paise coins can be placed in a line in \[\frac{(m+n)\,\,!}{m\,\,!\,\,n\,\,!}\] ways. If the extreme coins are ten paise coins, then the remaining \[n-2\] ten paise coins and \[m\] one rupee coins can be arragned in a line in \[\frac{(m+n-2)\,\,!}{m\,\,!(n-2)\,\,!}\] ways. Hence the required probability \[=\frac{\frac{(m+n-2)\,\,!}{m\,\,!(n-2)\,\,!}}{\frac{(m+n)\,\,!}{m\,\,!\,\,n\,\,!}}=\frac{n(n-1)}{(m+n)(m+n-1)}\].
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