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If 10 coins of Rs. 10, five coins of Rs...

If 10 coins of Rs. 10, five coins of Rs. 5 are to be placed in a line, then the probability that the extreme coins are of Rs. 5 is

A

`(1)/(15!)`

B

`(1)/(10!)`

C

`(5!10!)/(15!)`

D

None of these

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of finding the probability that the extreme coins are of Rs. 5 when placing 10 coins of Rs. 10 and 5 coins of Rs. 5 in a line, we can follow these steps: ### Step 1: Determine the total number of coins We have: - 10 coins of Rs. 10 - 5 coins of Rs. 5 Total coins = 10 + 5 = 15 coins. ### Step 2: Calculate the total arrangements of the coins The total arrangements of these 15 coins can be calculated using the formula for permutations of multiset: \[ \text{Total arrangements} = \frac{15!}{10! \times 5!} \] ### Step 3: Fix the extreme coins To find the favorable outcomes where both extreme coins are Rs. 5, we fix two Rs. 5 coins at the ends. This leaves us with: - 10 coins of Rs. 10 - 3 coins of Rs. 5 (since we used 2 Rs. 5 coins) Now we have a total of: 10 (Rs. 10) + 3 (Rs. 5) = 13 coins left to arrange in the middle. ### Step 4: Calculate the arrangements of the remaining coins The arrangements of these 13 coins can be calculated as: \[ \text{Favorable arrangements} = \frac{13!}{10! \times 3!} \] ### Step 5: Calculate the probability The probability that the extreme coins are Rs. 5 is given by the ratio of favorable arrangements to total arrangements: \[ P(\text{extreme coins are Rs. 5}) = \frac{\text{Favorable arrangements}}{\text{Total arrangements}} = \frac{\frac{13!}{10! \times 3!}}{\frac{15!}{10! \times 5!}} \] ### Step 6: Simplify the probability expression This simplifies to: \[ P = \frac{13! \times 5!}{15! \times 3!} \] Now, substituting \(15! = 15 \times 14 \times 13!\): \[ P = \frac{5!}{15 \times 14 \times 3!} \] ### Step 7: Calculate the factorials Calculating the factorials: - \(5! = 120\) - \(3! = 6\) Substituting these values: \[ P = \frac{120}{15 \times 14 \times 6} = \frac{120}{1260} = \frac{2}{21} \] ### Final Answer Thus, the probability that the extreme coins are of Rs. 5 is: \[ \frac{2}{21} \]
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