RowQ
The Vault
RowQ
The Vault
CBSE Class 11 Maths · 11 questions · 26 marks
Counting sounds easy until the list gets too long to write out — how many five-character codes, how many committees, how many arrangements of a word with repeated letters? This chapter gives you two tools that replace listing: permutations, which care about order, and combinations, which do not. The hardest skill here is not the formula but deciding which of the two a question is actually asking for.
The value of the permutation 8P3 is:
Answer
8P3 = 336. By definition nPr = n!/(n - r)!, so 8P3 = 8!/5! = 8 × 7 × 6. Multiplying, 8 × 7 = 56 and 56 × 6 = 336. The value 56 is 8C3 (order ignored) and 512 is 8³, which would count arrangements with repetition allowed.
The value of 10C8 is:
Answer
10C8 = 45. Use the symmetry nCr = nC(n - r), so 10C8 = 10C2, which is far quicker to compute. 10C2 = (10 × 9)/(2 × 1) = 90/2 = 45. The answer 90 is 10P2, which would be correct only if the order of selection mattered.
The number of distinct arrangements of all the letters of the word TRELLIS is:
Answer
There are 2520 distinct arrangements. The word has 7 letters: T, R, E, L, L, I, S, in which L occurs twice and every other letter once. For n letters with one letter repeated p times the count is n!/p!. So the number is 7!/2! = 5040/2 = 2520. The value 5040 forgets that swapping the two identical Ls does not create a new arrangement.
If nC2 = 28, then the value of n is:
Answer
n = 8. By definition nC2 = n(n - 1)/2, so n(n - 1)/2 = 28 and hence n(n - 1) = 56. We need two consecutive integers whose product is 56; since 8 × 7 = 56, we take n = 8. Formally, n² - n - 56 = 0 factorises as (n - 8)(n + 7) = 0, and n = -7 is rejected because n must be a positive integer.
Assertion (A): The number of ways of choosing a committee of 3 students from a group of 9 students is 84. Reason (R): When the order of selection is irrelevant, the number of ways of choosing r objects from n distinct objects is nCr = n!/(r!(n - r)!).
Answer
Both A and R are true and R is the correct explanation of A. Reason R is the correct definition of a combination. A committee is an unordered selection — swapping the names does not create a different committee — so the count is 9C3. 9C3 = (9 × 8 × 7)/(3 × 2 × 1) = 504/6 = 84, confirming Assertion A. Since the combination formula in R is exactly what yields 84, R correctly explains A.
How many 4-digit numbers can be formed using the digits 2, 3, 5, 7, 8 and 9 if no digit is repeated?
Answer
We are arranging 4 of the 6 available digits, and order matters because 2357 and 5723 are different numbers. So the count is 6P4 = 6!/(6 - 4)! = 6!/2!. This equals 6 × 5 × 4 × 3 = 360. Note that none of the digits is 0, so there is no restriction about a leading zero, and all 360 arrangements are valid 4-digit numbers.
From a group of 8 boys and 6 girls, in how many ways can a team of 5 be chosen so that it contains exactly 2 girls?
Answer
A team is an unordered selection, so combinations are used. Choose 2 girls from 6: 6C2 = (6 × 5)/(2 × 1) = 15 ways. The remaining 3 members must be boys, chosen from 8: 8C3 = (8 × 7 × 6)/(3 × 2 × 1) = 336/6 = 56 ways. By the fundamental principle of counting, the two choices are made together, so the total is 15 × 56 = 840. Hence there are 840 such teams.
In how many ways can 5 different books be arranged on a shelf if one particular book must always be placed at the leftmost position?
Answer
The leftmost position is fixed by the chosen book, so there is only 1 way to fill it. The remaining 4 distinct books can be arranged among the remaining 4 places in 4! ways. 4! = 4 × 3 × 2 × 1 = 24. So the total number of arrangements is 1 × 24 = 24.
Consider all the distinct arrangements of the letters of the word GEOLOGIST. (a) How many arrangements are there in total? (b) How many of them begin with the letter L? (c) How many of them have all four vowels together?
Answer
First analyse the word. GEOLOGIST has 9 letters: G, E, O, L, O, G, I, S, T. The letter G occurs twice, the letter O occurs twice, and E, L, I, S, T occur once each. (a) With 9 letters and two pairs of repeats, the number of distinct arrangements is 9!/(2! × 2!). 9! = 362880, and 2! × 2! = 4, so the total is 362880/4 = 90720. (b) Fix L in the first position; there is 1 way to do that. The remaining 8 letters are G, E, O, O, G, I, S, T, still containing two Gs and two Os. They can be arranged in 8!/(2! × 2!) = 40320/4 = 10080 ways. So 10080 arrangements begin with L. (c) The vowels are E, O, O and I — four vowels, with O repeated twice. Tie the four vowels into a single block. Then we arrange this block along with the five consonants G, L, G, S, T, i.e. 6 units in all, of which two (the Gs) are identical. Number of ways to arrange the units = 6!/2! = 720/2 = 360. Within the block the four vowels can be ordered in 4!/2! = 24/2 = 12 ways, dividing by 2! for the repeated O. By the multiplication principle the answer is 360 × 12 = 4320 arrangements.
A committee of 6 members is to be formed from 7 men and 5 women. (a) In how many ways can this be done if the committee must contain at least 3 women? (b) In how many ways can it be done if it must contain no woman? (c) Find the total number of possible committees and verify that your answer to (a) is consistent with it.
Answer
(a) 'At least 3 women' splits into three disjoint cases, since only 5 women are available. Case 1 — exactly 3 women and 3 men: 5C3 × 7C3 = 10 × 35 = 350. (5C3 = 5C2 = 10 and 7C3 = (7 × 6 × 5)/6 = 35.) Case 2 — exactly 4 women and 2 men: 5C4 × 7C2 = 5 × 21 = 105. Case 3 — exactly 5 women and 1 man: 5C5 × 7C1 = 1 × 7 = 7. Since the cases cannot overlap, add them: 350 + 105 + 7 = 462 committees. (b) With no woman, all 6 members are men chosen from 7: 7C6 = 7C1 = 7 committees. (c) The total number of committees of 6 from 12 people is 12C6. 12C6 = (12 × 11 × 10 × 9 × 8 × 7)/(6 × 5 × 4 × 3 × 2 × 1) = 665280/720 = 924. Consistency check: the remaining committees have at most 2 women, namely 0 women (7), 1 woman (5C1 × 7C5 = 5 × 21 = 105) and 2 women (5C2 × 7C4 = 10 × 35 = 350), totalling 7 + 105 + 350 = 462. Adding the two halves, 462 + 462 = 924 = 12C6, so the answer to (a) is consistent.
Read the following and answer the questions that follow: A sports club issues each member a five-character identity code. The first two characters are distinct letters chosen from the set {A, B, C, D, E, F}, and the last three characters are distinct digits chosen from 1 to 9. Order matters throughout, since AB123 and BA123 are different codes. (a) In how many ways can the two-letter part be formed? (b) In how many ways can the three-digit part be formed? (c) How many different identity codes are possible in all? (d) How many codes end with the digit 5?
Answer
(a) Two distinct letters are arranged from 6 available letters, so the count is 6P2 = 6 × 5 = 30 ways. (b) Three distinct digits are arranged from the 9 digits 1 to 9, so the count is 9P3 = 9 × 8 × 7 = 504 ways. (c) The letter part and the digit part are chosen independently, so by the fundamental principle of counting the total is 30 × 504 = 15120 codes. (d) Fix the last character as 5; there is 1 way. The other two digit positions must be filled by distinct digits from the remaining 8 digits, in 8P2 = 8 × 7 = 56 ways. The letter part is still free, in 30 ways. So the number of such codes is 30 × 56 × 1 = 1680.
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