07 ప్రస్తారణలు మరియు కలయికలు

వ్యాయామం 04

Question:

A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.

Answer:

Answer: Step 1: Calculate the total number of balls in the bag = 5 black balls + 6 red balls = 11 balls

Step 2: Calculate the total number of ways in which 2 black and 3 red balls can be selected = (5C2) × (6C3)

Step 3: Substitute the values in the formula = (5C2) × (6C3) = (5! / (2! × 3!)) × (6! / (3! × 3!))

Step 4: Simplify the expression = (5! / (2! × 3!)) × (6! / (3! × 3!)) = 10 × 20 = 200

Step 5: The number of ways in which 2 black and 3 red balls can be selected is 200.

Question:

Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.

Answer:

Answer: Step 1: Calculate the total number of balls = 6 red balls + 5 white balls + 5 blue balls = 16 balls

Step 2: Calculate the number of ways of selecting 3 balls of each colour = 6C3 x 5C3 x 5C3 = 120

Step 3: Calculate the number of ways of selecting 9 balls from 16 balls = 16C9 = 11,639,040

Step 4: Calculate the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls = 11,639,040 x 120 = 1,396,688,800

Question:

Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.

Answer:

  1. Count the number of aces in the deck of cards: 52 cards, 4 aces.

  2. Calculate the number of combinations of 5 cards out of 52 cards: 52C5 = 2,598,960.

  3. Multiply the number of combinations by the number of aces in the deck: 2,598,960 x 4 = 10,395,840.

  4. The number of 5 card combinations out of a deck of 52 cards with exactly one ace in each combination is 10,395,840.

Question:

In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?

Answer:

Step 1: Identify the total number of boys and girls. There are 5 boys and 4 girls.

Step 2: Calculate the total number of possible combinations. This can be calculated using the formula nCr, where n = total number of boys and girls and r = number of boys and girls to be selected.

In this case, n = 9 (5 boys + 4 girls) and r = 3 (3 boys + 3 girls).

Therefore, the total number of possible combinations is 9C3 = 84.

Question:

In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers?

Answer:

  1. There are 17 players from which to choose a cricket team of 11.
  2. Of the 11 players, exactly 4 must be bowlers.
  3. There are 5 players who can bowl, so there are 5 possible choices for the 4 bowlers.
  4. For each selection of 4 bowlers, there are (17-4=13) players left to choose from for the remaining 7 positions.
  5. There are (13 choose 7) ways to select the remaining 7 players from the 13.
  6. Therefore, the total number of ways to select a cricket team of 11 from 17 players in which only 5 players can bowl and each team must have exactly 4 bowlers is (5 x (13 choose 7)) = 5 x 1716 = 8580.

Question:

How many chords can be drawn through 21 points on a circle ? A 301 B 210 C 111 D 220

Answer:

Answer: D 220

Question:

In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?

Answer:

Step 1: Count the total number of courses available to the student, which is 9 courses.

Step 2: Subtract the number of compulsory courses (2) from the total number of courses available (9). This leaves the student with 7 courses to choose from.

Step 3: Calculate the number of possible combinations of 5 courses from the 7 available courses. This can be done using the formula for combinations, which is: nCr = n!/(r!(n-r)!).

Step 4: Plug the values into the formula. In this case, n = 7 and r = 5. This gives us 7!/(5!(7-5)!), which simplifies to 7!/5!.

Step 5: Calculate the result. 7!/5! = 7654321/54321 = 7*6 = 42.

Therefore, the student can choose a programme of 5 courses from the 9 courses available in 42 different ways.

JEE స్టడీ మెటీరియల్ (గణితం)

01 సెట్లు

02 సంబంధాలు మరియు విధులు

03 త్రికోణమితి విధులు

04 గణిత ప్రేరణ సూత్రం

05 సంక్లిష్ట సంఖ్యలు మరియు చతుర్భుజ సమీకరణాలు

06 లీనియర్ అసమానతలు

07 ప్రస్తారణలు మరియు కలయికలు

08 ద్విపద సిద్ధాంతం

09 సీక్వెన్సులు మరియు సిరీస్

10 స్ట్రెయిట్ లైన్స్ వ్యాయామం

10 స్ట్రెయిట్ లైన్స్ ఇతరాలు

11 కోనిక్ విభాగాలు

12 త్రీ డైమెన్షనల్ జామెట్రీకి పరిచయం

13 పరిమితులు మరియు ఉత్పన్నాలు

14 గణిత రీజనింగ్

15 గణాంకాలు

16 సంభావ్యత