GATE CSE 2000 | Question: 1.1

Question

The minimum number of cards to be dealt from an arbitrarily shuffled deck of $52$ cards to guarantee that three cards are from same suit is

• A. $3$
• B. $8$
• C. $9$
• D. $12$

Comments

This is a clear picture of 52 cards.

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The minimum # of pigeons which assures atleast K+1 pigeons in some pigeon hole=Kn+1

$Here\ K+1=3\ \&\ n=4(suits)$

$\therefore 2\times 4+1=9$

 

 

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In the deck of $52$ cards there are $4$ suits $[Heart\;\;Diamond\;\;Spade\;\;Club]$.

So, No. of holes = $4$

Now in each suit we pick $2$ cards. So, we picked $8$ cards and we will pick another card from same suit to guarantee that $3$ cards are from same suit.

$\therefore$ Min. no of cards = $8+1=9$

Answer 1

An easier way to think it as of worst case outcomes that is each time we take the card it is from different suit.

We know that there are 4 suits.

Let,

Heart= $H$

Spade= $S$

Diamond= $D$

Club= $C$

So the possible withdrawal sequence  (alternating/worst case)

 $HSDCHSDC\displaystyle$__ (this 9th card will make 3 cards of same suit.)

Therefore option C

Comments

nice one for understanding

Answer 2

Let Min Number of cards be N.

 

Given,

No of holes (holes) = 4 (since there are 4 suits in pack of 52 cards)

Required no of cards (R) = 3 (since 3 cards of same suit required)

Using Generalized Pigeon Hole,

$\left \lceil \frac{N}{holes}\right\rceil$ = R

$\left \lceil \frac{N}{4}\right\rceil$ = 3

So minimum N satisfying above equation is 9 ( since $\left \lceil \frac{9}{4}\right\rceil$ = 3).

Hence Option C is correct.