Aptitude Discussion

Q. |
Out of eight crew members three particular members can sit only on the left side. Another two particular members can sit only on the right side. Find the number of ways in which the crew can be arranged so that four men can sit on each side. |

✖ A. |
$864$ |

✖ B. |
$863$ |

✖ C. |
$865$ |

✔ D. |
$1728$ |

**Solution:**

Option(**D**) is correct

Required number of ways,

$= {^3C_2} × 4! ×4!$

$= \textbf{1728}$

**Fran**

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**Sushma Saroj**

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but nikhar one seat is needed by left side also and it can be selected from remaining 3 seats

**Chandrasekaran**

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Thank u nikar

**PRATYUSH ANAND**

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Simple way to solve it.

left side arrangement:- $^4P_3$

right side arrangement:- $^4P_2$

and remaining 3 members arrangement on 3 seats:- $^3P_3$

So Total ways:- $^4P_3*^4P_2*^3P_3=1728$

**PRATYUSH ANAND**

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Total 8 seats:-

4 seats left and 4 seats right

5 seats are fixed.

3 members on left + 2 members on right

Thus remaining (8-5)=3 members can seat on other remaining seats.

So,** left side arrangement:-**

Selecting 3 seats out of 4 seats and no. of ways 3 members can seats.

**4C3*3!=4P3 ----(i)**

1 seat remains on left side and select 1 member from 3 and the way he can seat.

**3C1*1!=3P1 ----(ii)**

Similarly, **right side arrangement:-**

Selecting 2 seats out of 4 seats and no. of ways 2 members can seats.

**4C2*2!=4P2 ----(iii)**

2 seats remain on right side and select 2 member from 2 and the way they can seat.

**2C2*2!=2P2 ----(iv)**

**Now combining (i),(ii),(iii)&(iv), we get**

**Total no. of ways = 4P3*3P1*4P2*2P2 = 1728**

**Nikhar**

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out of total; 8 places, (3 left +2 right) 5 are already selected .. left are 8-5=3 locations. out of which 2 have to be fitted on right side. so $^3c_2$ is performed ..

and 4 on each side ..

so 4! ways arrangement for each. so $^3c_2*4!*4!$

question is confusing. when they say 3 can sit on the left, do they mean the furthest left or any of the 4 seats on the left?