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6 months ago

Class 11th Permutations and Combinations

8. Is 3 ! + 4 ! = 7 ! ?

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Payal Gupta

Contributor-Level 10

8. L.H.S = 3! + 4!

= (1 * 2 * 3) + (1 * 2 * 3 * 4)

= 6 + 24

= 30

R.H.S = 7!

= 1 * 2 * 3 * 4 * 5 * 6 * 7

= 5040

As, L.H.S ≠ R.H.S

3! + 4! ≠ 7!

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New answer posted

6 months ago

Class 11th Permutations and Combinations

Permutation and Combinations Ex 6.2 Solutions

7. Evaluate

(i) 8 !

(ii) 4 ! – 3!

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Payal Gupta

Contributor-Level 10

1. We know that, n! = n (n – 1) (n – 2)…….

i. 8!

8! = 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8

= 40320

ii. 4! – 3!

= (1 * 2 * 3 * 4) – (1 * 2 * 3)

= 24 – 6

= 18

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6 months ago

Class 11th Probability

26. A die is thrown, find the probability of following events:

(i) A prime number will appear,

(ii) A number greater than or equal to 3 will appear,

(iii) A number less than or equal to one will appear,

(iv) A number more than 6 will appear,

(v) A number less than 6 will appear.

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alok kumar singh

Contributor-Level 10

26. The sample space of throwing s dice is

S = {1, 2, 3, 4, 5, 6}, n (S) = 6.

(i) Let A be event such that a prime number will appear. Then,

A = {2, 3, 5}

? n (A) = 3

Here; P (A) = $\frac{n (A)}{n (S)} = \frac{3}{6} = \frac{1}{2}$

(ii) Let B be event such that a number greater than or equal to 3 will appear. Then

B = {3, 4, 5, 6}

So, n (B) = 4

Therefore P (B) = $\frac{n (B)}{n (S)} = \frac{4}{6} = \frac{2}{3}$

(iii) Let C be event such that a number less than or equal to one will appear. Then,

C = {1}

So, n (C) = 1

? P (C) = $\frac{n (C)}{n (S)} = \frac{1}{6}$

(iv) Let D be event such that a number more than 6 appears. Then,

D =∅

So, n (D) = 0

? P (D) = $\frac{n (D)}{n (S)} = \frac{0}{6} = 0$

(v) Let E be event such that a number less than 6 appears. Then

E = {1, 2, 3, 4, 5}

...more

26. The sample space of throwing s dice is

S = {1, 2, 3, 4, 5, 6}, n (S) = 6.

(i) Let A be event such that a prime number will appear. Then,

A = {2, 3, 5}

? n (A) = 3

Here; P (A) = $\frac{n (A)}{n (S)} = \frac{3}{6} = \frac{1}{2}$

(ii) Let B be event such that a number greater than or equal to 3 will appear. Then

B = {3, 4, 5, 6}

So, n (B) = 4

Therefore P (B) = $\frac{n (B)}{n (S)} = \frac{4}{6} = \frac{2}{3}$

(iii) Let C be event such that a number less than or equal to one will appear. Then,

C = {1}

So, n (C) = 1

? P (C) = $\frac{n (C)}{n (S)} = \frac{1}{6}$

(iv) Let D be event such that a number more than 6 appears. Then,

D =∅

So, n (D) = 0

? P (D) = $\frac{n (D)}{n (S)} = \frac{0}{6} = 0$

(v) Let E be event such that a number less than 6 appears. Then

E = {1, 2, 3, 4, 5}

So, n (E) = 5

? P (E) = $\frac{n (E)}{n (S} = \frac{5}{6}$

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New answer posted

6 months ago

Class 11th Permutations and Combinations

6. Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?

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Payal Gupta

Contributor-Level 10

6. To generate a signal which requires 2 flags one below another we can have the following combination of any of the 5 flag at top and the one of the remaining 4 flag at the bottom.

Hence, total no. of possible combination = 5 * 4 = 20

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6 months ago

Class 11th Probability

25. A coin is tossed twice, what is the probability that atleast one tail occurs?

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alok kumar singh

Contributor-Level 10

25. When a coin is tossed twice we have the sample space

S = {TT, TH, HT, HH}

So, n (S) = 4

Let A be the event of getting at least one tail.

Then, A = {TH, HT, TT}

So, n (A) = 3

Therefore, P (A)= Numver of outcome favorable to A/Total possible outcomes = $\frac{}{} = \frac{n (A)}{n (S)}$ .

$= \frac{3}{4}$

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6 months ago

Class 11th Permutations and Combinations

5. A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?

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Payal Gupta

Contributor-Level 10

5. When a coin is tossed one time a head or a tail is the possible outcome.

So, when a coin is tossed 3 times the total number of possible outcomes = 2 * 2 * 2 = 8

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6 months ago

Class 11th Probability

24. Which of the following can not be valid assignment of probabilities for outcomes

of sample Space S = {ω₁, ω₂, ω₃, ω₄, ω₅, ω₆, ω₇,}

Assignment ω ₁ ω ₂ ω ₃ ω ₄ ω ₅ ω ₆ ω ₇

(a) 0.1 &nbs

...more

24. Which of the following can not be valid assignment of probabilities for outcomes

of sample Space S = {ω₁, ω₂, ω₃, ω₄, ω₅, ω₆, ω₇,}

Assignment ω ₁ ω ₂ ω ₃ ω ₄ ω ₅ ω ₆ ω ₇

(a) 0.1 0.01 0.05 0.03 0.01 0.2 0.6

(b)

(c) 0.1 0.2 0.3 0.4 0.5 0.6 0.7

(d) – 0.1 0.2 0.3 0.4 – 0.2 0.1 0.3

(e) $\frac{\bar{1}}{\bar{14}}$ $\frac{\bar{2}}{\bar{14}}$ $\frac{\bar{3}}{\bar{14}}$ $\frac{\bar{4}}{\bar{14}}$ $\frac{\bar{5}}{\bar{14}}$ $\frac{\bar{6}}{\bar{14}}$ $\frac{\bar{15}}{\bar{14}}$

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alok kumar singh

Contributor-Level 10

24. (a) P (S) = P (W1) + P (W2) +P (W3) + P (W4) +P (W5) +P (W6) +P (W7)

P (S) = 0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6

P (S) = 1

As the probability of sample space is 'one' the given assignment of probabilities is valid.

(b) P (S) = P (W1) + P (W2) +P (W3) + P (W4) +P (W5) +P (W6) +P (W7)

P (S) = $\frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7}$

$= \frac{1 + 1 + 1 + 1 + 1 + 1 + 1}{7} = \frac{7}{7} = 1$ .

P (S) = 1

Hence, the given assignment of probability is valid.

(c) P (S) = P (W1) + P (W2) +P (W3) + P (W4) +P (W5) +P (W6) +P (W7)

= 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.5 + 0.6 + 0.7

= 2.8

i.e., P (S) > 1

As probability of the sample space S should always be '1'. The given assignment is invalid.

(d) Here P (W1) = –0.1 is negative.

As p

...more

24. (a) P (S) = P (W1) + P (W2) +P (W3) + P (W4) +P (W5) +P (W6) +P (W7)

P (S) = 0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6

P (S) = 1

As the probability of sample space is 'one' the given assignment of probabilities is valid.

(b) P (S) = P (W1) + P (W2) +P (W3) + P (W4) +P (W5) +P (W6) +P (W7)

P (S) = $\frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7}$

$= \frac{1 + 1 + 1 + 1 + 1 + 1 + 1}{7} = \frac{7}{7} = 1$ .

P (S) = 1

Hence, the given assignment of probability is valid.

(c) P (S) = P (W1) + P (W2) +P (W3) + P (W4) +P (W5) +P (W6) +P (W7)

= 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.5 + 0.6 + 0.7

= 2.8

i.e., P (S) > 1

As probability of the sample space S should always be '1'. The given assignment is invalid.

(d) Here P (W1) = –0.1 is negative.

As probability of happening an event should always be positive and less than or equal to 1.

The given assignment is invalid.

(e) As P (W7) = $\frac{15}{14} = 1.07$

i.e. probability of happening an event is greater than one. The given assignment is invalid.

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New answer posted

6 months ago

Class 11th Permutations and Combinations

4. How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?

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Payal Gupta

Contributor-Level 10

4. For the 5-digit telephone number that can be constructed using 0 to 9 if each number starts with 67 and no digit appears more than once we can have

6 numbers

7 numbers

8 numbers

So total number of possible combination = 1 * 1 * 6 * 7 * 8 = 336

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New answer posted

6 months ago

Class 11th Permutations and Combinations

3. How many 4-letter code can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?

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Payal Gupta

Contributor-Level 10

3. To form a 4-letter code using the first 10 letters of the English alphabet without repeating we can have 10, 9, 8 and 7 numbers of letters to be filled at ones, tens, hundreds and thousands place simultaneously.

Hence, total no. of 4-letter code that can be made using the first 10-letter of English alphabet = 7 * 8 * 9 * 10 = 5040

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6 months ago

Class 11th Permutations and Combinations

2. How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?

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Payal Gupta

Contributor-Level 10

2. Since we are given with six numbers (1, 2, 3, 4, 5, 6) to form 3-digit even number and also repetition is allowed.

We can have only the number 2, 4, 6 in the ones place and all the six numbers can fill the tens and hundreds place.

So, total number of 3-digit even number that can be formed by 1, 2, 3, 4, 5, 6

= 6 * 6 * 3

= 108

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