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 .
Updated On: Jul 6, 2022
a
b
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The Correct Option isB
Solution and Explanation
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 $= \,^{7}C_{4} \times 1 = 35$
