A group of 8 applicants for a job includes 5 who speak Arabi

Question

8 graduates have applied for a job, including 5 who speak Arabic. How many subsets of these applicants include at least two who speak Arabic and at least one who does not?

(A) 168
(B) 182
(C) 192
(D) 200
(E) 208

nischaysetthi · Accepted Answer

5c2 *(3c1+3c2+3c3) = 10*(7) =70 
+ 5c3 *(3c1+3c2+3c3) = 10*(7) =70 
+ 5c4*(3c1+3c2+3c3) = 5*(7) =35 
+ 5c5*(3c1+3c2+3c3) = 1*(7) = 7

70+70+35+7= 182

chetan2u · Answer

Not likely to be seen on GMAT but good for improving your concepts on probability.

We can do the question in two ways..
1)  Total-invalid sets 
The number of ways to choose 0 or more items of given n items is 2^n, which is nothing but nC0+nC1+....nCn.
Thus total ways to select 0 or more graduates from 8 graduates is 2^8 or 256.

Now let us subtract invalid subsets..

a) when  none of Arabic speaking are chosen 
So, we choose from 3 non arabic - 2^3 or 8.

b) when  exactly one of Arabic speaking is chosen 
So, we choose any of the five Arabic speaking in 5C1 ways and groups from 3 non arabic in 2^3 or 8, totalling to 5C1*8 =40 ways.

c) when  none of non Arabic speaking are chosen 
So, we choose from 5 arabic - 2^5 or 32.

So total is 8+40+32 = 80 but we are   subtracting certain events twice.  
In (a) and (c) we have a case when none is chosen, so add one back.
In (b) and (c)..5C1*2^3 gives option when one Arabic speaking is chosen while non arabic could be 0 or 1...or 8. Thus thete will be cases where only Arabic speaking is chosen and there will be FIVE such cases. But when in (c) we choose subsets of only Arabic speaking, there will be groups of exactly one where one of Arabic speaking is selected. So subtract these 5 subsets..

Total 2^8-80+(1+5) =256-74 = 182.

2)  Choose valid sets..

a) Subtract ways when  none non arabic is chosen  = 2^3-1(case when none is chosen)

b) Subtract cases when  none or exactly one Arabic speaking is chosen 
None chosen is exactly one way.
Exactly one chosen is 5C1 or 5 ways.
So valid sets for Arabic speaking = 2^5-1-5 = 26

Total subsets is 7*26 = 182.

B

kevincan · Answer

I love your second method

Adit_ · Answer

Quite a few possibilities here:

Firstly we can fix number of non-arabic speaker as 1:
2,1
3,1
4,1
5,1 
Total possibilities = 30+30+15+3 =  78 .

Next we can keep one of arabic speakers fixed and move around non-arabic speaker:
2,2
2,3
This will be 10*3+10*1 =  40 .
3,2
3,3
This will again be  40 .
4,2
4,3
This will be 5*3+5*1 =  20 .
5,2
5,3
This will be 1*3+1*1 =  4 .

Total = 78+40+40+20+4 =  182 .

Answer:  Option B

____________________________________

Other way by removing bad cases but I am getting stuck somewhere, can any expert help me here?  chetan2u   kevincan

We have 5+3 as Arabic and Non- Arabic.
To have 1 Arabic, we need only Non-Arabic + 1 Arabic = 4 items:

= 2^4 - 1 = 15
Now we can choose 5 Arabians thus:
15*5 = 75.
But we also haveto deduct the subsets of non-arabic repeating 5 times, so subtract 4 times.
75 - (2^3 - 1)*4 = 75 - 28 = 47.

To have NO non-Arabic we need only Arabic:
A subset with only 5 items:
2^5 - 1 = 31.

Total possibilities in a subset = 8 items thus,
2^8 - 1 = 255.

There is also 5 cases used extra when counting the Arabic cases when used 5 at a time as well as when using 5C1 method.

Removng these cases I am getting:
255 - (47+31-5) =  182 .

EDIT: Got it! I realised that I subtracted 5 extra cases where Arabic by itself as a single entity is used. Thus I have to subtract 5 from the bad cases. I hope my approach is right, kindly let me know if there is a mistake in my approach, thankyou!

nischaysetthi · Answer

You can even do it from   TOTAL - NOT REQUIRED   approach

Total cases  --->  2^8 or 256

NOT req.  ---> 
1)ARABIC are 1 out of 5 AND Non arabic can be 0,1,2,3 = 5c1*2^3 = 40 
2)ARABIC are 0 out of 5 AND Non arabic can be 0,1,2,3 = 5c0*2^3 = 8

3)ARABIC can be 0,1,2,3,4,5 AND Non arabic are 0 out of 3 = 2^5*3c0 =  32 
 TOTAL NOT REQ.  =  80    ({But it includes some common cases)

COMMON CASES  --->
(5c1 *3c0) + (5c0*3c0) = 5+1= 6

ANSWER= TOTAL - NOT REQUIRED 
 = 2^8 - (80-6)
 = 256 - 74
 =   182 
 .

(COMMON CASES ARISE in point 3) ---when we did 2^5*3c0 --- The case 5c1*3c0 which was already considered in 1) and the case 5c0*3c0 which was already considered in 2)got considered again in point 3 so we reduced it)

A group of 8 applicants for a job includes 5 who speak Arabi

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A group of 8 applicants for a job includes 5 who speak Arabi

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