The number of ways of placing 18 mice in two experimental groups and one control group, with all group equally large, are

Question

The number of ways of placing 18 mice in two experimental groups and one control group, with all group equally large, are (A) (18 !/(16 !)3) (B) (18 !/6 ! 6 !) (C) (12 !/(6 !)2) (D) None of these

Tardigrade Admin · Accepted Answer

Number of ways of placing 18 mice in two experimental groups and one control group = 18 C6 × 12 C6 × 6 C6 =(18 !/12 ! 6 !) × (12 !/6 ! × 6 !) × 1 =(18 !/(6 !)3)

Q. The number of ways of placing 18 mice in two experimental groups and one control group, with all group equally large, are

$\frac{18 !}{( 16 ! ) ^{3}}$

$\frac{18 !}{6 ! 6 !}$

$\frac{12 !}{( 6 ! ) ^{2}}$

None of these

Number of ways of placing $18 mi ce$ in two experimental
groups and one control group $=^{18} C_{6} \times^{12} C_{6} \times^{6} C_{6}$
$= \frac{18 !}{12 ! 6 !} \times \frac{12 !}{6 ! \times 6 !} \times 1$
$= \frac{18 !}{( 6 ! ) ^{3}}$

Q. The number of ways of placing 18 mice in two experimental groups and one control group, with all group equally large, are

(16!)318!​

6!6!18!​

(6!)212!​

None of these

Number of ways of placing 18mice in two experimental groups and one control group =18C6​×12C6​×6C6​ =12!6!18!​×6!×6!12!​×1 =(6!)318!​

$\frac{18 !}{( 16 ! ) ^{3}}$

$\frac{18 !}{6 ! 6 !}$

$\frac{12 !}{( 6 ! ) ^{2}}$

Number of ways of placing $18 mi ce$ in two experimental
groups and one control group $=^{18} C_{6} \times^{12} C_{6} \times^{6} C_{6}$
$= \frac{18 !}{12 ! 6 !} \times \frac{12 !}{6 ! \times 6 !} \times 1$
$= \frac{18 !}{( 6 ! ) ^{3}}$