Given 5 different green dyes, four different blue dyes and three different red dyes, the number of combinations of dyes which can be chosen taking atleast one green and one blue dye is
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 $=\frac{6!}{2!} \times{ }^7 P_6 \times \frac{1}{3!} \times \frac{1}{2!}=151200$