Maths Solve the Following Question.(4 Marks)

(a) When the child is seated between two women.

5 men, 2 women, and a child are to be seated around a round table such that the child is seated between two women.

∴ the two women can be seated on either side of the child in 2! ways.

Let us consider these 3 (two women and a child) as one unit.

Now, this one unit is to be arranged with the remaining 5 men, i.e., a total of 6 units are to be arranged around a round table, which can be done in (6 – 1)! = 5! ways.

∴ Required number of arrangements = 5! × 2! = 120 × 2 = 240

B. Two men can be selected from 5 men in

${ }^5 C_2=\frac{5 !}{2 !(5-2) !}=\frac{5 \times 4 \times 3 !}{2 \times 3 !}=10$ ways.

Also, these two men can sit on either side of the child in 2! ways.

Let us take two men and a child as one unit.

Now, this one unit is to be arranged with the remaining 3 men and 2 women,

i.e., a total of 6 units (3 + 2 + 1) are to be arranged around a round table, which can be done in (6 – 1)! = 5! ways.

∴ Required number of arrangements = 10 × 2! × 5! = 10 × 2 × 120 = 2400

$\therefore$ Total number of words can be formed $=\frac{11 !}{4 ! 2 ! 2 !}$

When letters R and H are never together. Other than 2R, 2H there are 4A, 1S, 1T, 1M.

These letters can be arranged in $\frac{7 !}{4 !}$ ways $=210$.

These seven letters create 8 gaps in which 2R, 2H are to be arranged.

Number of ways to do $=\frac{{ }^8 P _4}{2 ! 2 !}=420$

Required number of arrangements = 210 × 420 = 88200.

When all vowels are together. There are 4 vowels in the word MAHARASHTRA, i.e., A, A, A, A.

Let us consider these 4 vowels as one unit, which can be arranged among themselves in $\frac{4 !}{4 !}$

= 1 way. This unit is to be arranged with 7 other letters in which ‘H’ is repeated 2 times, ‘R’ is repeated 2 times.

$\therefore$ Total number of arrangements $=\frac{8 !}{2 ! 2 !}$

B. When there is a girl at each end. 3 girls can be arranged at two ends in

${ }^3 P_2=\frac{3 !}{1 !}=3 \times 2=6$ ways.

And remaining 1 girl and 3 boys can be arranged between the two girls in ${ }^4 P _4=4 !=24$

ways. ∴ Required number of arrangements = 6 × 24 = 144

C. Boys and girls are at alternate places.

We can first arrange 3 girls among themselves in ${ }^3 P_3=3 !=6$ ways.

Let girls be denoted by G. G – G – G – There are 3 places marked by ‘-’ where 3 boys can be arranged in 3! = 6 ways. ∴ Total number of such arrangements = 6 × 6 = 36 OR Similarly, we can first arrange 3 boys in 3! = 6 ways and then arrange 3 girls alternately in 3! = 6 ways. ∴ Total number of such arrangements = 6 × 6 = 36 ∴ Required number of arrangements = 36 + 36 = 72

D. All boys sit together. Let us consider all boys as one group.

This one qroup with the other 3 girls can be arranged ${ }^4 P_4=4 !=24$ ways.

Also, 3 boys can be arranged among themselves in ${ }^3 P _3=3 !=6$ ways.

∴ Required number of arrangements = 24 × 6 = 144

(B)When no two vowels are together. There are 6 consonants in the word ALGORITHM

they can be arranged among themselves in ${ }^6 P _6=6 !=720$ ways.

Let consonants be denoted by C. _C _C_ C _C_C_C There are 7 places marked by ‘_’ in which 3 vowels can be arranged.

$\therefore$ Vowels can be arranged in ${ }^7 P_3=\frac{7 !}{(7-3) !}=\frac{7 \times 6 \times 5 \times 4 !}{4 !}=210$ ways.

∴ Required number of arrangements = 720 × 210 = 151200

(C)"When consonants are at even positions.

There are 4 even places and 6 consonants in the word ALGORITHM.

$\therefore 6$ consonants can be arranged at 4 even positions in $6 P 4=\frac{6 !}{(6-4) !}=\frac{6 \times 5 \times 4 \times 3 \times 2 !}{2 !}=360$

ways.

Remaining 5 letters ( 3 vowels and 2 consonants) can be arranged in odd position in ${ }^5 P_5=$

5! = 120 ways. ∴ Required number of arrangements = 360 × 120 = 43200"

(D)When O is the first and T is the last letter. All the letters of the word ALGORITHM are to be arranged among themselves such that arrangement begins with O and ends with T. ∴ Position of O and T are fixed.

$\therefore$ Other 7 letters can be arranged between $O$ and $T$ among themselves in ${ }^7 P_7=7 !=5040$

ways. ∴ Required number of arrangements = 5040