Maths Solve the Following Question.(2 Marks)

Number of teams with No doctor in any team $={ }^8 C _6$

$\therefore$ Required number of ways $={ }^{12} C _6-{ }^8 C _6$

= 924 – 28 = 896

If ${ }^n C_x={ }^n C_y$, then either $x=y$ or $x=n-y$

∴ 8 = 12 or 8 = n – 12

But 8 = 12 is not possible

∴ 8 = n – 12

∴ n = 20

Comparing on both sides, we get

$n =9$

$\begin{array}{ll}\therefore & 2 n=14 \\ \therefore & n=7\end{array}$

${ }^3 C_2=\frac{3 !}{2 !(3-2) !}=\frac{3 \times 2 !}{2 ! \times 1 !}=3$ ways.

Also, these two women can sit together in 2! ways.

Let us take two women as one unit.

Now, this one unit is to be arranged with the remaining 3 men and 1 woman,

i.e., a total of 5 units are to be arranged around a round table, which can be done in (5 – 1)! = 4! ways.

∴ Required number of arrangements = 3 × 2! × 4! = 3 × 2 × 24 = 144

$\therefore$ Required number of arrangements $=\frac{12 !}{2 ! 3 ! 2 !}$

When all the vowels,

i.e., ‘A’, ‘A’, ‘A’, ‘E’, ‘I’ are to be kept together.

Let us consider them as one unit.

Number of arrangements of these vowels among themselves $=\frac{5 !}{3 !}$ ways.

This unit is to be arranged with 7 other letters in which ‘M’ and ‘T’ repeated 2 times each.

$\therefore$ Number of such arrangements $=\frac{8 !}{2 ! 2 !}$

$\therefore$ Required number of arrangements $=\frac{8 ! \times 5 !}{2 ! 2 ! 3 !}$

done in ${ }^6 P _6=6 !=720$ ways.

(a) If the number is to be divisible by 5, the unit’s place digit can be 5 only. ∴ it can be arranged in 1 way only.

The other 5 digits can be arranged among themselves in ${ }^5 P _5=5 !=120$ ways.

∴ Required number of numbers divisible by 5 = 1 × 120 = 120

(B)If the number is not divisible by 5, unit’s place can be any digit from 3, 4, 6, 7, 8. ∴ it can be arranged in 5 ways.

Other 5 digits can be arranged in ${ }^5 P _5=5 !=120$ ways.

∴ Required number of numbers not divisible by 5 = 5 × 120 = 600

$\therefore 3$ vowels can be arranged at 3 even places in ${ }^3 P _3=3 !=6$ ways.

Also, 4 consonants can be arranged at 4 odd places in ${ }^4 P _4=4 !=24$ ways.

Required number of arrangements = 6 × 24 = 144

(b) When repetition of letters is not allowed, the number of 4-letter words formed from

the letters of the word MADHURI is ${ }^7 P _4=\frac{7 !}{(7-4) !}=\frac{7 \times 6 \times 5 \times 4 \times 3 !}{3 !}=840$

$\begin{aligned} \therefore \quad \frac{n !}{r !(n-r) !} & =\frac{15 !}{8 !(15-8) !} \\ & =\frac{15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 !}{8 ! \times 7 !} \\ & =\frac{15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} \\ & =5 \times 13 \times 11 \times 9 \\ & =6435\end{aligned}$

$\begin{aligned} \therefore \quad \frac{ n !}{ r !( n - r ) !} & =\frac{15 !}{10 !(15-10) !}=\frac{15 !}{10 ! \times 5 !} \\ & =\frac{15 \times 14 \times 13 \times 12 \times 11 \times 10 !}{10 ! \times(5 \times 4 \times 3 \times 2 \times 1)} \\ & =\frac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1} \\ & =7 \times 13 \times 3 \times 11 \\ & =3003\end{aligned}$