Aptitude Discussion

Q. |
How many arrangements of four 0's (zeroes), two 1's and two 2's are there in which the first 1 occur before the first 2? |

✖ A. |
420 |

✖ B. |
360 |

✖ C. |
320 |

✔ D. |
210 |

**Solution:**

Option(**D**) is correct

Total number of arrangements,

$= \dfrac{8!}{4!×2!×2!} = 420$

Since, there are two 1's and two 0's, the number of arrangements in which the first 1 is before the first 2 is same as the number of arrangement in which the first 2 is before the first 1 and they are each equal to half the total number of arrangements $= \textbf{210}$

**Numan**

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**Priyanka**

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why the 1st 1 before the first 2 is same as first 2 before the first 1

I think the answer is obtained by 7!/4!1!1! since you cant change the order of the first 1 and 2