Permutation-Combination
Aptitude

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Q.

For a Lofoya Assets, a selection committee is to be chosen consisting of 5 ex-technicians.

Now there are 12 representatives from four zones. It has further been decided that if Mr.  Lofs is selected, Relk and Lemini will not be selected and vice-versa.

In how many ways it can be done?

 A.

$572$

 B.

$672$

 C.

$472$

 D.

$372$

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Solution:
Option(B) is correct

${^{10}C_5}$:- when both are not included.

${^{10}C_4}$:- when one of them is included.

Number of ways,

$= {^{10}C_5} + {^{10}C_4} + {^{10}C_4}$

$= \textbf{672}$


(5) Comment(s)


EXC
 ()

WHAT IF WE REASON THIS WAY WE HAVE 12C5 WAYS WHEN IGNORING RESTRICTIONS AND FOR THE THREE TO BE SELECTED AT THE SAME TIME WE HAVE (9C2)(3C3) WAYS SUBTRACTING THESE FROM THE TOTAL WE HAVE 756?



Geek
 ()

Geek

Case 1: Lofs selected but R and Lem are not

9C4

Case 2: R & Lem selected but not Lofs

9C3

Case 3: All the 3 are not selected

9C5

Final answer = 9C4 + 9C3 + 9C5 = 336



Mahak
 ()

Two cases

1. when lofs is selected then

no of ways to select other 4

$9C4$

2.When lofs is not selected

$11C4$

total = $9C4+11C4=588$



Subra
 ()

Answer should be 588.

why 672 is wrong because, when you add 10C4 two times it will includes some combinations of Relk and Lemini's both the times so ans should be less than 672.

we have to take this problem with 5 different cases

1. Non of the three are in group which is 9C5.

2. Lof is IN the committee and Relk & Lemini are NOT ie. 9C4

3. Relk is IN the group and Lemini & Lof are NOT ie. 9C4

4. Lemini is IN and Lof & Relk are NOT. ie. 9C4

5. Relk & Lemini are IN but Lof isn't. ie. 9C3

9C5+9C4+9C4+9C4+9C3=588



Name
 ()

This is wrong... You're not even evaluating how many different ways the 3 guys can be selected when you do choose one of them. Furthermore, you say when "both aren't included", yet there are 3 of them.. Lofs, Relk, and Lemini.

Not choosing 1 of them:

= 9C5

Choosing 1 of them:

= (3C1 x 9C4)

= 9C5 + (3 x 9C4)

= 504