(B) To ensure that we have at least two balls of the same colour,we apply the Pigeonhole Principle.If we draw $1$ ball,it can be either black or white

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(B) To ensure that we have at least two balls of the same colour,we apply the Pigeonhole Principle.If we draw $1$ ball,it can be either black or white.If we draw $2$ balls,they could be one black and one white (different colours).If we draw $3$ balls,according to the Pigeonhole Principle,since there are only $2$ colours available,at least $2$ balls must be of the same colour.Therefore,the minimum number of balls required is $3$.

Competitive Quantitative Aptitude Permutation and Combination Permutation and Combination

Medium

$A$ bag contains balls of two colours,$3$ black and $3$ white. What is the minimum number of balls which must be drawn from the bag,without looking,so that among these there are two of the same colour?

A
$2$
B
$3$
C
$4$
D
$5$

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Two families with three members each and one family with four members are to be seated in a row. In how many ways can they be seated so that the same family members are not separated?

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The set $S = \{1, 2, 3, \ldots, 12\}$ is to be partitioned into three sets $A, B, C$ of equal size. Thus $A \cup B \cup C = S$ and $A \cap B = B \cap C = C \cap A = \emptyset$. The number of ways to partition $S$ is

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If the number of five-digit numbers with distinct digits and $2$ at the $10^{\text{th}}$ place is $336k$,then $k$ is equal to

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$A$ bag contains balls of two colours,$3$ black and $3$ white. What is the minimum number of balls which must be drawn from the bag,without looking,so that among these there are two of the same colour?

Similar Questions

Two families with three members each and one family with four members are to be seated in a row. In how many ways can they be seated so that the same family members are not separated?

If $33!$ is divisible by $2^n$,where $n \in N$,then the sum of all possible values of $n$ is:

The number of ways in which $3$ boys and $4$ girls can be seated around a circular table if no two boys sit together is:

The set $S = \{1, 2, 3, \ldots, 12\}$ is to be partitioned into three sets $A, B, C$ of equal size. Thus $A \cup B \cup C = S$ and $A \cap B = B \cap C = C \cap A = \emptyset$. The number of ways to partition $S$ is

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