Aptitude Discussion

Q. |
There are four hotels in a town. If 3 men check into the hotels in a day then what is the probability that each checks into a different hotel? |

✖ A. |
6/7 |

✖ B. |
1/8 |

✔ C. |
3/8 |

✖ D. |
5/9 |

**Solution:**

Option(**C**) is correct

Total cases of checking in the hotels $= 4^3$ ways.

Cases, when 3 men are checking in different hotels = $4×3×2=24$ ways.

Required probability:

\(=\dfrac{24}{4^3}\)

\(=\dfrac{3}{8}\)

**Edit:** Corrected the typo pointed out by**Bhargavi Bhat.**

**Edit:** For an alternative solution, check comment by **Ram Kr Prajapati.**

**ADARSH**

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**Ram Kr Prajapati**

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Total number of ways in which three people can visit hotel out of four hotel,

$={^4C_1} \times {^4C_1} \times {^4C_1}=64$.

Total number of ways in which each person visit different hotel,

$={^4C_1} \times {^3C_1} \times {^2C_1}=24$ (Since Whenever a person visits a hotel then another person is not allowed to visit this hotel).

So Required Probability,

$P=\dfrac{24}{64}$

$P=\dfrac{3}{8}$

good explanation..this is simple way to solve

**Bhargavi Bhat**

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instead of 43 ways it should be $4^3$ ways

Thank you Bhargavi for letting me know, corrected the mistake.

**Niranjana Murali**

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How did we arrive at 4 power 3 in total cases of cheking?

4C1*4C1*4C1*4C0=4*4*4=4^3=64