(D) The number of ways to select at least one green toy from $5$ is $2^5 - 1 = 31$ ways.The number of ways to select at least one blue toy from $4$ is

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(D) The number of ways to select at least one green toy from $5$ is $2^5 - 1 = 31$ ways.The number of ways to select at least one blue toy from $4$ is $2^4 - 1 = 15$ ways.The number of ways to select any number of red toys from $3$ (including zero) is $2^3 = 8$ ways.Since these selections are independent,the total number of combinations is $31 \times 15 \times 8$.

Class 11 Mathematics Permutation and Combination Definition of combinations, Condition combinations

Medium

Given $5$ different green toys,$4$ different blue toys,and $3$ different red toys,how many combinations of toys can be chosen taking at least one green and one blue toy?

AP EAMCET 2022 All AP EAMCET

A
$32 \times 16 \times 4$
B
$31 \times 15 \times 4$
C
$32 \times 16 \times 8$
D
$31 \times 15 \times 8$

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Permutation and Combination

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Definition of combinations, Condition combinations

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Given $5$ different green toys,$4$ different blue toys,and $3$ different red toys,how many combinations of toys can be chosen taking at least one green and one blue toy?

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Statement-$1:$ The number of ways of distributing $10$ identical balls in $4$ distinct boxes such that no box is empty is $^9C_3$.Statement-$2:$ The number of ways of choosing any $3$ places from $9$ different places is $^9C_3$.

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Statement-$1:$ The number of ways of distributing $10$ identical balls in $4$ distinct boxes such that no box is empty is $^9C_3$.
Statement-$2:$ The number of ways of choosing any $3$ places from $9$ different places is $^9C_3$.