Permutation And Combination - Study Mode

[#131] Ten different letters of alphabet are given, words with 5 letters are formed from these given letters. Then, the number of words which have at least one letter repeated is:
Correct Answer

(A) 69760

Explanation

Solution: Number of words which have at least one letter replaced, = Total number of words - total number of words in which no letter is repeated. = 10 5 – 16 P 5 = 100000 − 30240 = 69760

[#132] 12 chairs are arranged in a row and are numbered 1 to 12. 4 men have to be seated in these chairs so that the chairs numbered 1 to 8 should be occupied and no two men occupy adjacent chairs. Find the number of ways the task can be done.
Correct Answer

(B) 384

Explanation

Solution: Given there are 12 numbered chairs, such that chairs numbered 1 to 8 should be occupied. 1, 2, 3, 4, 5, 6, 7, 8 , 9, 10, 11, 12 The various combinations of chairs that ensure that no two men are sitting together are listed. (1, 3, 5, __ ), The fourth chair can be 5,6,10,11 or 12, hence 5 ways. (1, 4, 8, __ ), The fourth chair can be 6,10,11 or 12 hence 4 ways. (1, 5, 8, __ ), the fourth chair can be 10,11 or 12 hence 3 ways. (1, 6, 8, __ ), the fourth chair can be 10,11 or 12 hence 3 ways. (1, 8, 10, 12) is also one of the combinations. Hence, 16 such combinations exist. In case of each these combinations we can make the four men inter arrange in 4! ways. Hence, the required result =16 × 4! = 384

[#133] How many words can be formed by re-arranging the letters of the word ASCENT such that A and T occupy the first and last position respectively?
Correct Answer

(C) 4!

Explanation

Solution: As A and T should occupy the first and last position, the first and last position can be filled in only one following way. A _ _ _ _ T. The remaining 4 positions can be filled in 4! Ways by the remaining words (S, C, E, N, T). Hence by rearranging the letters of the word ASCENT we can form, 1 x 4! = 4! Words.

[#134] If 6 P r = 360 and If 6 C r = 15, find r ?
Correct Answer

(C) 4

Explanation

Solution: n P r = n C r × r! 6 P r = 15 × r! 360 = 15 × r! r! = $$frac{{360}}{{15}}$$ = 24 r! = 4 × 3 × 2 × 1 ⇒ r! = 4! Therefore, r = 4

[#135] In how many ways can six different rings be worn on four fingers of one hand?
Correct Answer

(D) 4096

Explanation

Solution: Each ring may be worn in any of 4 fingers. So, each ring may be worn in 4 different ways. ∴ 6 rings may be worm in (4×4×4×4×4×4) = 4 6 = 4096 ways.