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11. i. n = 6, r = 2

$\frac{6!}{\left(6-2\right)!}$

= $\frac{6!}{4!}$

= $\frac{6\times 5\times \left(4!\right)}{\left(4!\right)}$

= 30

ii. n = 9, r = 5

$\frac{9!}{\left(9-5\right)!}$

= $\frac{9!}{4!}$

= $\frac{9\times 8\times 7\times 6\times 5\times \left(4!\right)}{4!}$

= 9 × 8 × 7 × 6 × 5

= 15,120

29. Number of women in the city council n (A) = 6

As there are four men and six women the total number of person in the sample space is 4 + 6 = 10.

So, n (S) = 10

P (A) = $\frac{n(A)}{n(S)}=\frac{6}{10}=\frac{3}{5}$

Yes, it is possible to take admission into the Bharathiar University Distance Education BA course without appearing for an entrance exam. The admissions to the BA course are made based on marks obtained in the Class 12 examinations. Moreover, those who are offered admission are required to pay the first installment of the fees.

ICRI Pune admit candidates to the MBA programme on entrance score basis. Candidates who meet the eligibility criteria have to first submit the applications. The applicants then have to appear for the ICRI Written Exam. Candidates shortlisted based on exam scores are further called for the personal interview round. Once final selection is done, aspirants can pay the admission fee amount to confirm their seats. 

10. We have,

$\frac{1}{6!}$ + $\frac{1}{7!}$ = $\frac{x}{8!}$

=> $\frac{1}{6!}$ + $\frac{1}{7*6!}$ = $\frac{x}{8 *7*6!}$

=> 1 + $\frac{1}{7}$ = $\frac{x}{8*7}$

=> $\frac{8}{7}$ = $\frac{x}{8 *7}$

=>x = 8 * 8

=>x = 64

9.  $\frac{8!}{6! * 2!}$ = $\frac{8 *7*\left(6!\right) }{\left(6!\right)*1*2}$ = 4 * 7 = 28

28. The sample space of the experiment is

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

So, n (S) = 12.

(i) Let E be event such that sum of numbers that turn up is 3. Then,

E = { (1, 2)}

So, n (E) = 1

P (E) = $\frac{n(E)}{n(S)}=\frac{1}{12}$ .

(ii) Let F be event such that sum of number than turn up is 12. Then,

F = { (6, 6)}

So, n (F) = 1

P (F) = $\frac{n(F)}{n(S)}=\frac{1}{12}$ .

Students seeking admission to the Bharathiar University Distance Education BA programme must begin with an eligibility check. As per the eligibility criteria, students who have completed Class 12 or an equivalent examination from a recognised board of India are eligible to apply. Students must also note that the aforementioned criteria is taken from official sources. However, it is subject to change.

27. (a) Since there are 52 cards in the sample space,

n (S) = 52.

So, there are 52 sample points.

(b) In a deck of 52 cards there are 4 ace cards of which only one is of spades.

Hence, if A be an event of getting an ace of spades.

n (A) = 1

So, P (A) = $\frac{n(A)}{n(S)}=\frac{1}{52}$ .

(c) (i) Let B be an event of drawing an ace. As there are 4 ace cards we have,

n (B) = 4

So, P (B) = $\frac{n(B)}{n(S)}=\frac{4}{52}=\frac{1}{13}$ .

(ii) Let D be an event of drawing black cards. Since there are 26 black cards we have,

n (D) = 26.

So, P (D) = $\frac{n(D)}{n(S)}=\frac{26}{52}=\frac{1}{2}$

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!

Candidates can download the official UPSC ESE (IES) prelims answer key by the methods provided below:

• Visit the official website of UPSC – upsc.gov.in
• Click on the 'Examination' tab given at the top
• A drop list will appear, click on the Answer key link appearing on the list
• The ESE answer key PDF link will appear in the list
• Click on the link and get the ESE cutoff for all the categories and phases

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

Once the Bharathiar University Distance Education BA selection process is completed, selected students must pay the tuition fees to secure their admission. As per official sources, the total tuition fee to pursue BA courses ranges from INR 21,100.

Note: The above-mentioned fee is as per the official sources. However, it is indicative and subject to change.

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

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 = $\frac{}{}=\frac{n(A)}{n(S)}$ .

$=\frac{3}{4}$