GATE Electrical 2020 | GA | Question: 5

Question

If $\text{P, Q, R, S}$ are four individuals, how many teams of size exceeding one can be formed, with $\text{Q}$ as a member?

• A. $5$
• B. $6$
• C. $7$
• D. $8$

Comments

we need group of 4 exceeding 1 already including Q,

there will a total of 8 as 2^3 but we need more than 1 ,so 8 -1 =7

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Answer Here!   

Answer 1

Ans C = 7.

Number of Individuals $= 4,$ team Size greater than $1,$ Q is premium member. 
_ _ _ Q. (max three positions left to be filled and three members left: P, R, S)

By raw counting we can get $7: \{PQ, RQ,SQ, PRQ, PSQ, RSQ, PQRS\}.$

Alternatively we can do as follows:

As four individuals are defined distinct, and we just have to select them for the three remaining positions.

Selecting $1$ person out of three, selecting $2$ persons out of three and selecting all $3$ persons, we get

$^{3}\textrm{C}_{1} + ^{3}\textrm{C}_{2} + ^{3}\textrm{C}_{3}$

$=3 + 3 + 1 = 7$

Comments

@Abhrajyoti00 , @JAINchiNMay

why can’t i do it this way?

as teams can be of size 2,3,4.

so for a team of 2 →   _ Q → the first position can be filled in 3 ways, either (P,R or S) and Q fixed.

for a team of 3 → _ _ Q → first position 3 ways and 2^nd position 2 ways.

for a team of 4 → _ _ _ Q → only 1 way.

EDIT: got it, I was counting same team twice in team size 3.

Answer 2

Ans : C

--> 2^3 -1

Exp:

Q is their already in team,

now we have 3 people left

Total Team possibility = 8

But of the Team will have only Q,

But acc to question , team must have atleast 2 people,So we need to subtract by 1

So final ans = 8-1 = 7

Answer 3

Team size = 4 => PQRS

Team size = 3 => PQR, QPS, QRS

Team size = 2 =>  QP, QR, QS

Total = 7