Number of ways in which 5 alike red balls and 6 alike green balls can be arranged so that exactly two pairs of green balls are together and separated from each other, is

Question

Number of ways in which 5 alike red balls and 6 alike green balls can be arranged so that exactly two pairs of green balls are together and separated from each other, is (A) 45 (B) 90 (C) 144 (D) 15

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/85972730045e43df2bba20702fc4c164-.png" /> selection of 2 more gaps from the remaining 4 in 4 C2 ways for G and G ∴ Total ways = 6 C 2 ⋅ 4 C 2=(6 !/2 ! 4 !) ⋅ (4 !/2 ! 2 !)=(6 !/2 ! 2 ! 2 !)=(720/8)=90.

Q. Number of ways in which 5 alike red balls and 6 alike green balls can be arranged so that exactly two pairs of green balls are together and separated from each other, is

45

90

144

15

$&$ selection of 2 more gaps from the remaining 4 in $^{4} C_{2}$ ways for $G$ and $G$
$∴$ Total ways $=^{6} C_{2} \cdot^{4} C_{2} = \frac{6 !}{2 ! 4 !} \cdot \frac{4 !}{2 ! 2 !} = \frac{6 !}{2 ! 2 ! 2 !} = \frac{720}{8} = 90$ .

Q. Number of ways in which 5 alike red balls and 6 alike green balls can be arranged so that exactly two pairs of green balls are together and separated from each other, is

45

90

144

15

& selection of 2 more gaps from the remaining 4 in 4C2​ ways for G and G ∴ Total ways =6C2​⋅4C2​=2!4!6!​⋅2!2!4!​=2!2!2!6!​=8720​=90.

$&$ selection of 2 more gaps from the remaining 4 in $^{4} C_{2}$ ways for $G$ and $G$
$∴$ Total ways $=^{6} C_{2} \cdot^{4} C_{2} = \frac{6 !}{2 ! 4 !} \cdot \frac{4 !}{2 ! 2 !} = \frac{6 !}{2 ! 2 ! 2 !} = \frac{720}{8} = 90$ .