Question

Fill in the blanks. (i) The number of different words that can be formed from the letters of the word such that two vowels never come together is . (ii) Three balls are drawn from a bag containing $5$ red, $4$ white and $3$ black balls. The number of ways in which this can be done if atleast $2$ are red is . (iii) The total number of ways in which six $'+'$ and four $'-'$ signs can be arranged in a line such that no two signs $'-'$ occur together, is .

• A. a
• B. b
• C. c
• D. d

Answer 1

Answer: (B)

b

Answer 2

Total number of letters in the word $= 12$ Number of consonants $= 6$, Number of vowels $= 6$ When we fix consonants at six places then there are seven places for vowels as shown below. $6$ consonants out of which $2$ are alike can be placed in $\frac{6!}{2!}$ ways and $6$ vowels, out of which $3\, E??$ alike and $2\, I's$ can be arranged at seven places in $^{7}P_{6} \times\frac{1}{3!} \times \frac{1}{2!}$ ways. $\therefore$ Total number of words $= \frac{6!}{2!} \times\,^{7}P_{6} \times\frac{1}{3!} \times \frac{1}{2!} = 151200$ (ii) Required number of ways $= \left(^{5}C_{2} \times\,^{7}C_{1}\right) + \,^{5}C_{3} = \left(10 \times 7\right)+10$ $=70 + 10$ $= 80$ (iii) The arrangement of signs is shown in the following figure. Thus, $'+'$ sign can be arranged in $1$ way because all are identical and $4$ negative signs can be arranged at $7$ places . in $^{7}C_{4}$ ways. $\therefore$ Total number of ways $= \,^{7}C_{4} \times 1 = 35$