If ${ }^n P_r=840$ and ${ }^n C_r=35$, then $r$ is equal to ............ .
Given that, ${ }^n P_r=840$ and ${ }^n C_r=35$
$$\begin{array}{ll} \because\quad { }^n P_r={ }^n C_r \cdot r! \\ \Rightarrow \quad &840=35 \times r! \\ \Rightarrow & r!=\frac{840}{35}=24 \\ \Rightarrow & r!=4 \times 3 \times 2 \times 1 \\ \Rightarrow & r!=4! \\ \Rightarrow & r=4 \end{array}$$
${ }^{15} C_8+{ }^{15} C_9-{ }^{15} C_6-{ }^{15} C_7$ is equal to ............ .
$$\begin{aligned} { }^{15} C_8+{ }^{15} C_9-{ }^{15} C_6-{ }^{15} C_7 & ={ }^{15} C_{15-8}+{ }^{15} C_{15-9}-{ }^{15} C_6-{ }^{15} C_7 \quad\left[\because{ }^n C_r={ }^n C_{n-r}\right] \\ & ={ }^{15} C_7+{ }^{15} C_6-{ }^{15} C_6-{ }^{15} C_7 \\ & =0 \end{aligned} $$
The number of permutations of n different objects, taken r at a line, when repetitions are allowed, is ............ .
Number of permutations of n different things taken r at a time when repetition is allowed $=n^r$
The number of different words that can be formed from the letters of the word 'INTERMEDIATE' such that two vowels never come together is ........... .
Total number of letters in the word 'INTERMEDIATE' $=12$ out of which 6 are consonants and 6 are vowels. The arrangement of these 12 alphabets in which two vowels never come together can be understand with the help of follow manner.
| V | C | V | C | V | C | V | C | V | C | V | C | V |
6 consonants out of which 2 are alike can be placed in $\frac{6!}{2!}$ ways and 6 vowels, out of which 3 E's alike and 2 I's are alike can be arranged at seven place in ${ }^7 P_6 \times \frac{1}{3!} \times \frac{1}{2!}$ ways.
$\therefore$ Total number of words $=\frac{6!}{2!} \times{ }^7 P_6 \times \frac{1}{3!} \times \frac{1}{2!}=151200$
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.
$$\begin{aligned} \text { Required number of ways } & ={ }^5 \mathrm{C}_2 \times{ }^7 \mathrm{C}_1+{ }^5 \mathrm{C}_3 \quad \text{[since, at least two red]}\\ & =10 \times 7+10 \\ & =70+10=80 \end{aligned}$$