We may have [3 men and 2 women] or [4 men and 1 woman] or [5 men only].
| $\therefore$ Required number of ways | = (7C3 x 6C2) + (7C4 x 6C1) + (7C5) |
| =$ \left(\dfrac{7 \times 6 \times 5}{3 \times 2 \times 1} \times\dfrac{6 \times 5}{2 \times 1} \right) $+ (7C3 x 6C1) + (7C2) | |
| = 525 +$ \left(\dfrac{7 \times 6 \times 5}{3 \times 2 \times 1} \times 6\right) $+$ \left(\dfrac{7 \times 6}{2 \times 1} \right) $ | |
| = (525 + 210 + 21) | |
| = 756. |
The word OPTICAL contains 7 different letters.
When the vowels OIA are always together, they can be supposed to form one letter.
Then, we have to arrange the letters PTCL [OIA].
Now, 5 letters can be arranged in 5! = 120 ways.
The vowels [OIA] can be arranged among themselves in 3! = 6 ways.
$\therefore$ Required number of ways = $\left(120 \times 6\right)$ = 720.
We may have [1 boy and 3 girls] or [2 boys and 2 girls] or [3 boys and 1 girl] or [4 boys].
| $\therefore$ Required number of ways | = (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4) |
| = (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2) | |
| =$\left (6 \times 4\right)$ +$ \left(\dfrac{6 \times 5}{2 \times 1} \times\dfrac{4 \times 3}{2 \times 1} \right) $+$ \left(\dfrac{6 \times 5 \times 4}{3 \times 2 \times 1} \times 4\right) $+$ \left(\dfrac{6 \times 5}{2 \times 1} \right) $ | |
| = (24 + 90 + 80 + 15) | |
| = 209. |
5 subjects can be arranged in 6 periods in 6P5 ways.
Any of the 5 subjects can be organized in the remaining period 5C1 ways.
Two subjects are alike in each of the arrangement. So we need to divide by 2! to avoid overcounting.
Total number of arrangements
$=\dfrac{~^{6}P_5 × ~^5C_1}{2!}=1800$
| Required number of ways = (7C5 x 3C2) = (7C2 x 3C1) =$ \left(\dfrac{7 \times 6}{2 \times 1} \times 3\right) $= 63. |
There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants.
Let us mark these positions as under:
(1) (2) (3) (4) (5) (6)
Now, 3 vowels can be placed at any of the three places out 4, marked 1, 3, 5.
Number of ways of arranging the vowels = 3P3 = 3! = 6.
Also, the 3 consonants can be arranged at the remaining 3 positions.
Number of ways of these arrangements = 3P3 = 3! = 6.
Total number of ways = $\left(6 \times 6\right)$ = 36.
Number of ways of selecting [3 consonants out of 7] and [2 vowels out of 4]
| = (7C3 x 4C2) | |
| =$ \left(\dfrac{7 \times 6 \times 5}{3 \times 2 \times 1} \times\dfrac{4 \times 3}{2 \times 1} \right) $ | |
| = 210. |
Number of groups, each having 3 consonants and 2 vowels = 210.
Each group contains 5 letters.
| Number of ways of arranging 5 letters among themselves= 5!= 5 x 4 x 3 x 2 x 1= 120. |
$\therefore$ Required number of ways = $\left(210 \times 120\right)$ = 25200.
The word LEADING has 7 different letters.
When the vowels EAI are always together, they can be supposed to form one letter.
Then, we have to arrange the letters LNDG [EAI].
Now, 5 [4 + 1 = 5] letters can be arranged in 5! = 120 ways.
The vowels [EAI] can be arranged among themselves in 3! = 6 ways.
$\therefore$ Required number of ways = $\left(120 \times 6\right)$ = 720.
Around a circle, 5 boys can be arranged in 4! ways.
Given that the boys and the girls alternate. Hence there are 5 places for the girls. Therefore the girls can be arranged in 5! ways.
Total number of ways
=4!×5!=24×120=2880The word BIHAR has 5 letters and all these 5 letters are different.
Total number of words that can be formed by using all these 5 letters
= 5P5=5
=5×4×3×2×1=120