Solve the Following Question.(5 Marks)

Question 15 Marks

A committee of 10 persons is to be formed from a group of 10 women and 8 men. How many possible committees will have at least 5 women? How many possible committees will have men in majority?

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Question 25 Marks

If ${ }^n C_{r-1}=6435,{ }^n C_r=5005,{ }^n C_{r+1}=3003$, find ${ }^r C_5$.

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Question 35 Marks

Find the number of permutations of letters in each of the following words.
(i) DIVYA
(II)SHANTARAM
(III)REPRESENT
(IV)COMBINE
(V)BAL BHARATI

Answer

I. There are 5 distinct letters in the word DIVYA. ∴ Number of permutations of the letters of the word DIVYA = 5! = 120II. There are 9 letters in the word SHANTARAM in which ‘A’ is repeated 3 times.
$\therefore$ Number of permutations of the letters of the word SHANTARAM $=\frac{9 !}{3 !}$
= 9 × 8 × 7 × 6 × 5 × 4 = 60480
III. There are 9 letters in the word REPRESENT in which ‘E’ is repeated 3 times and ‘R’ is repeated 2 times.$\therefore$ Number of permutations of the letters of the word REPRESENT $=\frac{9 !}{3 ! 2 !}$
$=\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4}{2}$
$=30240$
IV. There are 7 distinct letters in the word COMBINE. ∴ Number of permutations of the letters of the word COMBINE = 7! = 5040
V. There are 10 letters in the word BALBHARATI in which ‘B’ is repeated 2 times and ‘A’ is repeated 3 times$\therefore$ Number of permutations of the letters of the word BALBHARATI $=\frac{10 !}{213 !}$
$=\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 3 \times 2}$
$=302400$

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Question 45 Marks

Find the number of ways so that letters of the word HISTORY can be arranged as

(A) Y and T are together

(B)Y is next to T

(C)there is no restriction

(D)begin and end with a vowel

(E)end in ST

(F)begin with S and end with T

Answer

There are 7 letters in the word HISTORY (a) When ‘Y’ and ‘T’ are together. Let us consider ‘Y’ and ‘T’ as one unit. This unit with other 5 letters are to be arranged.

$\therefore$ The number of arrangement of one unit and 5 letters $={ }^6 P_6=6 !=720$

Also, ' $Y$ ' and ' $T$ ' can be arranged among themselves in ${ }^2 P _2=2 !=2$ ways.

∴ A total number of arrangements when Y and T are always together = 6! × 2! = 120 × 2 = 1440

When ‘Y’ is next to ‘T’. Let us take this (‘Y’ next to ‘T’) as one unit. This unit with 5 other letters is to be arranged.

$\therefore$ The number of arrangements of 5 letters and one unit $={ }^6 P _6=6 !=720$

Also, ‘Y’ has to be always next to ‘T’. ∴ They can be arranged among themselves in 1 way only. ∴ Total number of arrangements possible when Y is next to T = 720 × 1 = 720

When there is no restriction.

7 letters can be arranged among themselves in ${ }^7 P _7=7$ ! ways.

∴ The total number of arrangements possible if there is no restriction = 7!

When begin and end with a vowel. There are 2 vowels in the word HISTORY. All other letters of the word HISTORY are to be arranged between 2 vowels such that the arrangement begins and ends with a vowel.The other 5 letters can be filled between the two vowels in ${ }^5 P_5=5 !=120$ ways.

Also, 2 vowels can be arranged among themselves at first and last places in ${ }^2 P _2=2 !=2$

ways. ∴ Total number of arrangements when the word begins and ends with vowel = 120 × 2 = 240

When a word ends in ST. As the arrangement ends with ST,

the remaining 5 letters can be arranged among themselves in ${ }^5 P_5=5 !=120$ ways.

∴ Total number of arrangements when the word ends with ST = 120

When a word begins with S and ends with T. As arrangement begins with S and ends with T,the remaining 5 letters can be arranged between $S$ and $T$ among themselves in ${ }^5 P_5=5 !=$

120 ways. Total number of arrangements when the word begins with S and ends with T = 120

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Maths Solve the Following Question.(5 Marks)

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