07 ଅନୁମତି ଏବଂ ମିଶ୍ରଣ
ବ୍ୟାୟାମ 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:
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Count the number of aces in the deck of cards: 52 cards, 4 aces.
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Calculate the number of combinations of 5 cards out of 52 cards: 52C5 = 2,598,960.
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Multiply the number of combinations by the number of aces in the deck: 2,598,960 x 4 = 10,395,840.
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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:
- There are 17 players from which to choose a cricket team of 11.
- Of the 11 players, exactly 4 must be bowlers.
- There are 5 players who can bowl, so there are 5 possible choices for the 4 bowlers.
- For each selection of 4 bowlers, there are (17-4=13) players left to choose from for the remaining 7 positions.
- There are (13 choose 7) ways to select the remaining 7 players from the 13.
- 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 ଅଧ୍ୟୟନ ସାମଗ୍ରୀ (ଗଣିତ)
01 ସେଟ୍
02 ସମ୍ପର୍କ ଏବଂ କାର୍ଯ୍ୟ
03 ଟ୍ରାଇଗୋନେଟ୍ରିକ୍ କାର୍ଯ୍ୟଗୁଡ଼ିକ
04 ଗାଣିତିକ ଅନୁକରଣର ନୀତି
05 ଜଟିଳ ସଂଖ୍ୟା ଏବଂ ଚତୁର୍ଭୁଜ ସମୀକରଣ
06 ରେଖା ଅସମାନତା
07 ଅନୁମତି ଏବଂ ମିଶ୍ରଣ
08 ଦ୍ୱିପାକ୍ଷିକ ତତ୍ତ୍।
09 କ୍ରମ ଏବଂ କ୍ରମ
10 ସିଧା ଲାଇନ ବ୍ୟାୟାମ
10 ସିଧା ରେଖା ବିବିଧ
11 କନିକ୍ ବିଭାଗ
12 ତିନୋଟି ଡାଇମେନ୍ସନାଲ୍ ଜ୍ୟାମିତିର ପରିଚୟ
13 ସୀମା ଏବଂ ଡେରିଭେଟିଭ୍
14 ଗାଣିତିକ କାରଣ
15 ପରିସଂଖ୍ୟାନ
16 ସମ୍ଭାବନା