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8. Permutations
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Q6 of 140 Page 288

If 15Pr–1 : 16Pr–2, = 3 : 4, find r.

To find: the value of r


Formula Used:


Total number of ways in which n objects can be arranged in r places (Such that no object is replaced) is given by,


nPr


15Pr–1 : 16Pr–2, = 3 : 4











Since r cannot be 21 as it creates a negative factorial in denominator. Therefore,
is not possible.


Hence, value of r is 14


More from this chapter

All 140 →
4

(i) If If 5Pr = 2 × 6Pr–1, find r.

(ii) If 20Pr = 13 × 20Pr–1, find r.


(iii) If 11Pr = 12Pr–1, find r.


5

(i) If nP4 : nP5, = 1 : 2, find n.

(ii) If n–1P3 : n+1P3, = 5 : 12, find n.


7

If 2n–1Pn : 2n+1Pn–1, = 22 : 7, find n.

8

Find n, If n+5Pn+1 = (n – 1). n+3Pn, find n.

Questions · 140
8. Permutations
1 2 3 4 5 6 7 8 9 10 11 12 13 14 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 1 1 1 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 1 2 3 4 5 6 7 1 2 3 4 5 6 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
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